Horizontal Projectile Motion Questions in English

(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
For the first ball,$H_1 = \frac{u^2 \sin^2 \theta}{2g}$.
For the second ball,$H_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
Given $u = 40\,m/s$ and $g = 10\,m/s^2$,we have $H_1 = \frac{1600 \sin^2 \theta}{20} = 80 \sin^2 \theta$ and $H_2 = 80 \cos^2 \theta$.
Given $H_2 - H_1 = 50\,m$,so $80(\cos^2 \theta - \sin^2 \theta) = 50$.
Using $\cos^2 \theta - \sin^2 \theta = \cos(2\theta)$,we get $80 \cos(2\theta) = 50$,so $\cos(2\theta) = \frac{5}{8} = 0.625$.
Since $\cos(2\theta) = 1 - 2\sin^2 \theta$,then $2\sin^2 \theta = 1 - 0.625 = 0.375$,so $\sin^2 \theta = 0.1875$.
Thus,$H_1 = 80 \times 0.1875 = 15\,m$.
Then $H_2 = H_1 + 50 = 15 + 50 = 65\,m$.

(B) The maximum height $H$ of a projectile is given by the formula $H = \frac{u^2 \sin^2 \alpha}{2g}$,where $u$ is the initial speed and $\alpha$ is the angle of projection with the horizontal.
For the first body,the angle with the horizontal is $\alpha_1 = \theta$. Thus,$H_1 = \frac{u^2 \sin^2 \theta}{2g}$.
For the second body,the angle with the vertical is $\theta$,which means the angle with the horizontal is $\alpha_2 = 90^\circ - \theta$. Thus,$H_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
The ratio of the maximum heights is $\frac{H_1}{H_2} = \frac{u^2 \sin^2 \theta / 2g}{u^2 \cos^2 \theta / 2g} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$.
Therefore,the ratio is $\tan^2 \theta : 1$.

(A) The instantaneous velocity is given by $v = a \hat{i} + (b - ct) \hat{j}$.
Comparing this with the standard projectile motion equations $v_x = u_x$ and $v_y = u_y - gt$,we get:
Initial horizontal velocity,$u_x = a$
Initial vertical velocity,$u_y = b$
Acceleration due to gravity,$g = c$
For a projectile,the time of flight $T$ is the time when the vertical displacement is zero,or when $v_y$ changes from $b$ to $-b$. Setting $v_y = 0$ at the peak,$b - ct = 0 \implies t = b/c$. Thus,the total time of flight $T = 2t = 2b/c$.
The horizontal range $R$ is given by $R = u_x \times T$.
Substituting the values,$R = a \times (2b/c) = 2ab/c$.

(B) The total horizontal range $R$ is the distance from the point of projection to the point where the ball falls. Given the wall is at $x = 6 \, m$ and the ball falls $18 \, m$ beyond the wall,the total range is $R = 6 + 18 = 24 \, m$.
The equation of the trajectory is given by $y = x \tan \theta \left(1 - \frac{x}{R}\right)$.
Here,$y = 3 \, m$ at $x = 6 \, m$,and $R = 24 \, m$.
Substituting these values into the equation:
$3 = 6 \tan \theta \left(1 - \frac{6}{24}\right)$
$3 = 6 \tan \theta \left(1 - \frac{1}{4}\right)$
$3 = 6 \tan \theta \left(\frac{3}{4}\right)$
$3 = \frac{18}{4} \tan \theta$
$3 = 4.5 \tan \theta$
$\tan \theta = \frac{3}{4.5} = \frac{30}{45} = \frac{2}{3}$
Therefore,$\theta = \tan^{-1}\left(\frac{2}{3}\right)$.

(B) The time of flight $T$ for a projectile is given by $T = \frac{2 u \sin \theta}{g}$.
Given $T = 2 \, s$,we have $2 = \frac{2 u \sin \theta}{g}$,which simplifies to $u \sin \theta = g$.
The maximum height $H$ attained by the projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting $u \sin \theta = g$ into the equation for $H$,we get $H = \frac{(g)^2}{2g} = \frac{g}{2}$.
Taking $g = 10 \, m/s^2$,we find $H = \frac{10}{2} = 5 \, m$.

(D) The vertical motion of the projectile is independent of the horizontal motion. Since the vertical acceleration remains $-g$,the maximum height $H$ remains unchanged.
For the horizontal range,the initial range is $R = u_x T$,where $T = \frac{2 u_y}{g}$ is the time of flight.
The new horizontal range $R^{\prime}$ with horizontal acceleration $a_x = \frac{g}{4}$ is given by:
$R^{\prime} = u_x T + \frac{1}{2} a_x T^2$
Substituting $R = u_x T$ and $T^2 = \frac{4 u_y^2}{g^2}$:
$R^{\prime} = R + \frac{1}{2} \left(\frac{g}{4}\right) \left(\frac{4 u_y^2}{g^2}\right)$
Since $H = \frac{u_y^2}{2g}$,we have $u_y^2 = 2gH$. Substituting this:
$R^{\prime} = R + \frac{g}{8} \left(\frac{4(2gH)}{g^2}\right) = R + \frac{g}{8} \left(\frac{8gH}{g^2}\right) = R + H$
Thus,the new range is $(R+H)$ and the height is $H$.

(B) Let the vertical component of velocity be $v_y = u \cos \theta$ (since the angle is with the vertical).
The equation for vertical displacement is $h = v_y t - \frac{1}{2} g t^2$.
For the two poles at times $t_1 = 1\,s$ and $t_2 = 3\,s$:
$h = v_y(1) - \frac{1}{2} g(1)^2$ ... $(1)$
$h = v_y(3) - \frac{1}{2} g(3)^2$ ... $(2)$
Equating $(1)$ and $(2)$: $v_y - 0.5g = 3v_y - 4.5g$.
$2v_y = 4g \implies v_y = 2g = 2 \times 9.8 = 19.6\,m/s$.
Now,find $h$ using $(1)$: $h = 19.6(1) - 4.9(1)^2 = 14.7\,m$.
However,the question asks for the maximum height $H_{max}$ relative to the launch point.
$H_{max} = \frac{v_y^2}{2g} = \frac{(19.6)^2}{2 \times 9.8} = \frac{19.6 \times 19.6}{19.6} = 19.6\,m$.

(C) Given that the horizontal range $R$ is four times the maximum height $H$,i.e.,$R = 4H$.
The formula for horizontal range is $R = \frac{U^2 \sin(2\theta)}{g} = \frac{2U^2 \sin\theta \cos\theta}{g}$.
The formula for maximum height is $H = \frac{U^2 \sin^2\theta}{2g}$.
Substituting these into the given condition: $\frac{2U^2 \sin\theta \cos\theta}{g} = 4 \times \frac{U^2 \sin^2\theta}{2g}$.
Simplifying the equation: $2 \sin\theta \cos\theta = 2 \sin^2\theta$.
Dividing both sides by $2 \sin\theta$ (assuming $\sin\theta \neq 0$): $\cos\theta = \sin\theta$.
Therefore,$\tan\theta = 1$,which implies $\theta = 45^{\circ}$.

(C) The horizontal component of velocity remains constant throughout the motion.
$u_x = u \cos 60^{\circ} = 150 \times \frac{1}{2} = 75\, ms^{-1}$.
Let the velocity at time $t$ be $v$. At this time,the angle with the horizontal is $45^{\circ}$.
The horizontal component of velocity at time $t$ is $v_x = v \cos 45^{\circ}$.
Since $v_x = u_x$,we have $v \cos 45^{\circ} = 75$,so $v = \frac{75}{\cos 45^{\circ}} = 75\sqrt{2}\, ms^{-1}$.
The vertical component of velocity at time $t$ is $v_y = v \sin 45^{\circ} = 75\sqrt{2} \times \frac{1}{\sqrt{2}} = 75\, ms^{-1}$.
Using the equation $v_y = u_y - gt$,where $u_y = u \sin 60^{\circ} = 150 \times \frac{\sqrt{3}}{2} = 75\sqrt{3}\, ms^{-1}$:
$75 = 75\sqrt{3} - 10t$.
$10t = 75(\sqrt{3} - 1)$.
$t = 7.5(\sqrt{3} - 1)\,s$.

(A) The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
We need to find the ratio $\frac{H}{T^2}$.
$\frac{H}{T^2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{(\frac{2u \sin \theta}{g})^2} = \frac{u^2 \sin^2 \theta}{2g} \times \frac{g^2}{4u^2 \sin^2 \theta}$.
Simplifying the expression,we get $\frac{H}{T^2} = \frac{g}{8}$.
Taking $g = 10 \ m/s^2$,the ratio is $\frac{10}{8} = \frac{5}{4}$.

(A) Given:
Initial horizontal velocity $u_x = 10\,m/s$.
Initial vertical velocity $u_y = 0\,m/s$.
Acceleration due to gravity $g = 10\,m/s^2$.
Time $t = 1\,s$.
$(A)$ Horizontal component of velocity $(v_x)$: In projectile motion,horizontal velocity remains constant. So,$v_x = u_x = 10\,m/s$. Matches with $(q)$.
$(B)$ Vertical component of velocity $(v_y)$: Using $v_y = u_y + g_y t$,where $g_y = g = 10\,m/s^2$. So,$v_y = 0 + 10(1) = 10\,m/s$. Matches with $(q)$.
$(C)$ Horizontal displacement $(x)$: $x = u_x t = 10(1) = 10\,m$. Matches with $(q)$.
$(D)$ Vertical displacement $(y)$: $y = u_y t + \frac{1}{2} g t^2 = 0(1) + \frac{1}{2}(10)(1)^2 = 5\,m$. Matches with $(p)$.
Therefore,the correct matching is $(A \rightarrow q, B \rightarrow q, C \rightarrow q, D \rightarrow p)$.

(D) For the ball to return to the boy's hand,the net displacement of the ball must be zero. This implies that the path of the ball must be a straight line relative to the boy.
This is possible only if the initial velocity vector and the net acceleration vector are collinear (acting along the same line).
Let the initial velocity be $\vec{v}_0$ at an angle $\theta$ with the vertical.
The horizontal component of acceleration is $a_x = 4\, m/s^2$ and the vertical component is $a_y = g = 10\, m/s^2$ (downwards).
The net acceleration vector $\vec{a}$ makes an angle $\alpha$ with the vertical,where $\tan \alpha = \frac{a_x}{a_y} = \frac{4}{10} = 0.4$.
Since the initial velocity must be in the same direction as the acceleration for the ball to return to the starting point,the angle $\theta$ with the vertical must be equal to $\alpha$.
Therefore,$\theta = \tan^{-1} (0.4)$.

(D) The angular momentum $L$ about the point of projection is given by $L = m(\vec{r} \times \vec{v})$.
At the highest point, the velocity is $v = u \cos \theta$ (horizontal direction).
The perpendicular distance from the point of projection to the velocity vector is the maximum height $H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$.
Thus, $L = m(u \cos \theta) \times H_{\max}$.
Substituting $H_{\max}$, we get $L = m(u \cos \theta) \left( \frac{u^2 \sin^2 \theta}{2g} \right) = \frac{m u^3 \cos \theta \sin^2 \theta}{2g}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$, we can write $\cos \theta \sin \theta = \frac{\sin 2\theta}{2}$.
Substituting this into the expression: $L = \frac{m u^3 \sin \theta}{2g} \left( \frac{\sin 2\theta}{2} \right) = \frac{m u^3 \sin \theta \sin 2\theta}{4g}$.

(B) The horizontal range $R$ of a projectile is given by: $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The maximum height $H$ of a projectile is given by: $H = \frac{u^2 \sin^2 \theta}{2g}$.
Dividing $H$ by $R$:
$\frac{H}{R} = \frac{u^2 \sin^2 \theta}{2g} \times \frac{g}{2u^2 \sin \theta \cos \theta}$.
Simplifying the expression:
$\frac{H}{R} = \frac{\sin \theta}{4 \cos \theta} = \frac{1}{4} \tan \theta$.
Therefore,$\frac{R}{H} = \frac{4}{\tan \theta} = 4 \cot \theta$.

(D) The formula for the horizontal range $R$ of a projectile is given by:
$R = \frac{u^2 \sin(2\theta)}{g}$
where $u$ is the initial velocity,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
If the initial velocity is doubled,the new velocity $u' = 2u$.
The new range $R'$ is:
$R' = \frac{(2u)^2 \sin(2\theta)}{g} = \frac{4u^2 \sin(2\theta)}{g}$
$R' = 4 \times \left( \frac{u^2 \sin(2\theta)}{g} \right) = 4R$
Therefore,the range becomes four times the original range.

(A) The maximum height $H$ attained by a projectile thrown with initial velocity $u$ at an angle $\theta = 90^{\circ}$ is given by:
$H = \frac{u^2}{2g}$
From this,we can express the square of the initial velocity as:
$u^2 = 2gH$
For maximum horizontal range $R_{\max}$,the projectile must be thrown at an angle $\theta = 45^{\circ}$. The formula for maximum horizontal range is:
$R_{\max} = \frac{u^2}{g}$
Substituting the value of $u^2$ from the height equation:
$R_{\max} = \frac{2gH}{g} = 2H$
Therefore,the maximum horizontal distance is $2H$.

(A) The maximum height $h$ for a projectile is given by $h = \frac{u^2 \sin^2 \theta}{2g}$.
For two angles of projection $\theta$ and $(90^\circ - \theta)$,the range $R$ is the same,given by $R = \frac{u^2 \sin 2\theta}{g}$.
Let $h_1 = \frac{u^2 \sin^2 \theta}{2g}$ and $h_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
Multiplying $h_1$ and $h_2$:
$h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \left( \frac{u^2 \cos^2 \theta}{2g} \right) = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}$.
Since $R = \frac{u^2 (2 \sin \theta \cos \theta)}{g}$,we have $R^2 = \frac{u^4 (4 \sin^2 \theta \cos^2 \theta)}{g^2}$.
Thus,$h_1 h_2 = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2} = \frac{R^2}{16}$.
Therefore,$R^2 = 16 h_1 h_2$,which implies $R = 4 \sqrt{h_1 h_2}$.

(A) When a bomb is released from a horizontally flying aeroplane,it possesses the same horizontal velocity as the aeroplane at the moment of release.
Due to gravity,it experiences a constant downward acceleration $(g)$.
The horizontal motion is uniform (constant velocity),while the vertical motion is uniformly accelerated.
The combination of these two independent motions results in a path that is a parabola.

(B) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$,where $u$ is the initial speed and $\theta$ is the angle of projection.
For the first projectile: $\theta_1 = 60^\circ$. Thus,$R_1 = \frac{u^2 \sin(2 \times 60^\circ)}{g} = \frac{u^2 \sin(120^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \sqrt{3}}{2g}$.
For the second projectile: $\theta_2 = 30^\circ$. Thus,$R_2 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \sqrt{3}}{2g}$.
Since $\sin(2 \times 60^\circ) = \sin(120^\circ) = \sin(60^\circ) = \sin(2 \times 30^\circ)$,the ranges are equal when the sum of the angles is $90^\circ$ (complementary angles). Therefore,their range will be same.

(C) For the person to catch the ball,the horizontal component of the ball's velocity must be equal to the constant speed of the person.
This ensures that the ball remains directly above the person throughout the flight,meaning the ball's motion relative to the person is purely vertical.
Equating the horizontal velocity of the ball to the speed of the person:
$v_0 \cos \theta = \frac{v_0}{2}$
Dividing both sides by $v_0$:
$\cos \theta = \frac{1}{2}$
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = 60^o$.
Thus,the person will be able to catch the ball if the angle of projection is $60^o$.

(D) The boy throws the ball from the roof of a building of height $10\, m$. We need to find the horizontal distance from the throwing point when the ball is at a height of $10\, m$ from the ground.
Since the ball is thrown from a height of $10\, m$ and we are looking for the distance when it is again at a height of $10\, m$,this corresponds to the horizontal range of the projectile on a horizontal plane passing through the point of projection.
The formula for the horizontal range $R$ is given by:
$R = \frac{u^2 \sin(2\theta)}{g}$
Given:
Initial speed $u = 10\, m/s$
Angle of projection $\theta = 30^o$
Acceleration due to gravity $g = 10\, m/s^2$
Substituting the values:
$R = \frac{(10)^2 \sin(2 \times 30^o)}{10} = \frac{100 \times \sin(60^o)}{10} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\, m$
Thus,the ball will be at a distance of $5\sqrt{3}\, m$ from the throwing point.

(A) The horizontal distance between the two buildings is $d = 100 \; m$. The bullets are fired towards each other with a horizontal velocity $v = 25 \; m/s$ each.
The relative horizontal velocity is $v_{rel} = v_1 + v_2 = 25 + 25 = 50 \; m/s$.
The time taken for the bullets to collide is $t = \frac{d}{v_{rel}} = \frac{100}{50} = 2 \; s$.
In this time,both bullets fall vertically due to gravity. The vertical displacement $s_y$ is given by $s_y = -\frac{1}{2} gt^2$.
Substituting the values: $s_y = -\frac{1}{2} \times 10 \times (2)^2 = -5 \times 4 = -20 \; m$.
The height of the collision from the ground is $H_{collision} = H_{initial} + s_y = 200 - 20 = 180 \; m$.
Thus,the bullets collide after $2 \; s$ at a height of $180 \; m$.

(N/A) The horizontal range $R$ of a projectile launched with initial velocity $v_{o}$ at an angle $\theta_{o}$ is given by the formula:
$R = \frac{v_{o}^{2} \sin(2\theta_{o})}{g}$
Consider two angles of projection,$\theta_{1} = 45^{\circ} + \alpha$ and $\theta_{2} = 45^{\circ} - \alpha$,where $\alpha$ is the equal amount by which they exceed or fall short of $45^{\circ}$.
For $\theta_{1} = 45^{\circ} + \alpha$,the range $R_{1}$ is:
$R_{1} = \frac{v_{o}^{2} \sin(2(45^{\circ} + \alpha))}{g} = \frac{v_{o}^{2} \sin(90^{\circ} + 2\alpha)}{g} = \frac{v_{o}^{2} \cos(2\alpha)}{g}$
For $\theta_{2} = 45^{\circ} - \alpha$,the range $R_{2}$ is:
$R_{2} = \frac{v_{o}^{2} \sin(2(45^{\circ} - \alpha))}{g} = \frac{v_{o}^{2} \sin(90^{\circ} - 2\alpha)}{g} = \frac{v_{o}^{2} \cos(2\alpha)}{g}$
Since $R_{1} = R_{2}$,it is proven that for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal.

(N/A) The maximum height is given by $h_{m} = \frac{(v_{0} \sin \theta_{0})^{2}}{2g} = \frac{(28 \sin 30^{\circ})^{2}}{2 \times 9.8} = \frac{14 \times 14}{19.6} = 10.0 \; m$.
$(b)$ The time taken to return to the same level (time of flight) is $T_{f} = \frac{2 v_{0} \sin \theta_{0}}{g} = \frac{2 \times 28 \times \sin 30^{\circ}}{9.8} = \frac{28}{9.8} \approx 2.86 \; s$ (rounded to $2.9 \; s$).
$(c)$ The horizontal range is $R = \frac{v_{0}^{2} \sin 2\theta_{0}}{g} = \frac{28^{2} \times \sin 60^{\circ}}{9.8} = \frac{784 \times 0.866}{9.8} \approx 69.28 \; m$ (rounded to $69 \; m$).

(A) Given: Speed of the ball,$u = 40\; m/s$. Maximum height,$h = 25\; m$. Acceleration due to gravity,$g = 9.8\; m/s^2$.
In projectile motion,the maximum height reached by a body projected at an angle $\theta$ is given by the relation:
$h = \frac{u^2 \sin^2 \theta}{2g}$
Substituting the values:
$25 = \frac{(40)^2 \sin^2 \theta}{2 \times 9.8}$
$25 = \frac{1600 \sin^2 \theta}{19.6}$
$\sin^2 \theta = \frac{25 \times 19.6}{1600} = 0.30625$
$\sin \theta = \sqrt{0.30625} = 0.5534$
$\theta = \sin^{-1}(0.5534) = 33.60^{\circ}$
The horizontal range $R$ is given by:
$R = \frac{u^2 \sin(2\theta)}{g}$
$R = \frac{(40)^2 \sin(2 \times 33.60^{\circ})}{9.8}$
$R = \frac{1600 \times \sin(67.2^{\circ})}{9.8}$
$R = \frac{1600 \times 0.9219}{9.8} \approx 150.53\; m$.

(B) The maximum horizontal range $R$ is given by $R = \frac{u^2}{g}$,where $u$ is the initial velocity and $g$ is the acceleration due to gravity. Given $R = 100 \; m$,we have $\frac{u^2}{g} = 100 \; m$.
To achieve the maximum vertical height $H$,the ball must be thrown vertically upward at an angle of $90^{\circ}$.
The formula for maximum vertical height is $H = \frac{u^2}{2g}$.
Substituting the value of $\frac{u^2}{g} = 100 \; m$ into the height formula:
$H = \frac{1}{2} \times \left( \frac{u^2}{g} \right) = \frac{1}{2} \times 100 \; m = 50 \; m$.
Thus,the cricketer can throw the ball to a maximum height of $50 \; m$.

(N/A) At the extreme position,the velocity of the bob is zero. If the string is cut at this moment,the bob will fall vertically downward under the influence of gravity. The trajectory is a straight line.
$(b)$ At the mean position,the velocity of the bob is $1\; m s^{-1}$ in the horizontal direction. If the string is cut at this moment,the bob will have an initial horizontal velocity and will be subject to gravity. Therefore,it will follow a parabolic trajectory.

(N/A) Projectile Motion: When an object is thrown into the Earth's gravitational field,it moves with a constant horizontal velocity and a constant vertical acceleration. Such a two-dimensional motion is called projectile motion,and the object is called a projectile.
For example,when we kick a football,it performs projectile motion if air resistance is neglected.
$A$ projectile has motion along a horizontal path with uniform velocity $(a_x = 0)$.
$A$ projectile has motion along a vertical path under gravity with uniform acceleration equal to $g$ $(a_y = -g)$.
Suppose the projectile is launched with velocity $\vec{v}_0$ at an angle $\theta_0$ with the $X$-axis. The acceleration is $\vec{a} = -g \hat{j}$.
The initial velocity components are:
$v_{0x} = v_0 \cos \theta_0$
$v_{0y} = v_0 \sin \theta_0$
Using the kinematic equation $r = r_0 + v_0 t + \frac{1}{2} a t^2$:
For the $X$-coordinate:
$x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 = 0 + (v_0 \cos \theta_0) t + 0 = v_0 \cos \theta_0 t$
For the $Y$-coordinate:
$y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 = 0 + (v_0 \sin \theta_0) t - \frac{1}{2} g t^2 = v_0 \sin \theta_0 t - \frac{1}{2} g t^2$
Thus,the position of the particle at any instant $t$ is given by the coordinates $(x(t), y(t)) = (v_0 \cos \theta_0 t, v_0 \sin \theta_0 t - \frac{1}{2} g t^2)$.

(N/A) Definition: An object projected into the air,which then moves under the influence of gravity alone,is called a projectile,and its motion is called projectile motion.
Derivation:
Let an object be projected with an initial velocity $v_0$ at an angle $\theta_0$ with the horizontal. The components of initial velocity are $v_{0x} = v_0 \cos \theta_0$ and $v_{0y} = v_0 \sin \theta_0$.
The horizontal position at time $t$ is given by:
$x = (v_0 \cos \theta_0)t \implies t = \frac{x}{v_0 \cos \theta_0} \quad \dots(1)$
The vertical position at time $t$ is given by:
$y = (v_0 \sin \theta_0)t - \frac{1}{2}gt^2 \quad \dots(2)$
Substituting the value of $t$ from equation $(1)$ into equation $(2)$:
$y = (v_0 \sin \theta_0) \left( \frac{x}{v_0 \cos \theta_0} \right) - \frac{1}{2}g \left( \frac{x}{v_0 \cos \theta_0} \right)^2$
Simplifying the expression:
$y = x \tan \theta_0 - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$
This is the equation of the trajectory of a projectile.

(N/A) The range $(R)$ is defined as the horizontal distance traveled by a projectile from its initial position $(x=0, y=0)$ to the final position $(x=R, y=0)$ where it returns to the same horizontal level.
At any instant of time $t$,the horizontal position is given by $x = (v_0 \cos \theta_0) t$.
At the time of flight $t_F$,the horizontal distance is $R = (v_0 \cos \theta_0) t_F$.
Since the time of flight is $t_F = \frac{2 v_0 \sin \theta_0}{g}$,we substitute this into the range equation:
$R = (v_0 \cos \theta_0) \left( \frac{2 v_0 \sin \theta_0}{g} \right)$
$R = \frac{v_0^2 (2 \sin \theta_0 \cos \theta_0)}{g}$
Using the trigonometric identity $\sin 2\theta_0 = 2 \sin \theta_0 \cos \theta_0$,we get:
$R = \frac{v_0^2 \sin 2\theta_0}{g}$
For maximum range $(R_{\max})$,the value of $\sin 2\theta_0$ must be maximum,which is $1$.
$\sin 2\theta_0 = 1 \implies 2\theta_0 = 90^\circ \implies \theta_0 = 45^\circ$.
Substituting $\sin 2\theta_0 = 1$ into the range formula:
$R_{\max} = \frac{v_0^2}{g}$.

(N/A) Projectile motion is the form of motion experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the action of gravity only. The path followed by a projectile is called its trajectory.
$A$ projectile particle is any object that is given an initial velocity and then allowed to move in space under the influence of gravity alone. Examples include a ball thrown in the air,a bullet fired from a gun,or a javelin thrown by an athlete.

(N/A) The range $R$ of a projectile is given by the formula: $R = \frac{v_{0}^2 \sin(2\theta_{0})}{g}$,where $v_{0}$ is the initial velocity,$\theta_{0}$ is the angle of projection,and $g$ is the acceleration due to gravity.
At the maximum height,the vertical component of the velocity becomes zero $(v_{y} = 0)$.
Therefore,the velocity of the projectile at the maximum height is equal to its horizontal component: $v = v_{x} = v_{0} \cos \theta_{0}$.

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\theta)}{g}$,where $u$ is the initial velocity,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
To maximize the range $R$,the value of $\sin(2\theta)$ must be maximum.
The maximum value of $\sin(2\theta)$ is $1$,which occurs when $2\theta = 90^{\circ}$.
Therefore,$\theta = 45^{\circ}$.
Thus,a particle should be projected at an angle of $45^{\circ}$ to cover the maximum range.

(N/A) The horizontal range $R$ of a projectile launched with initial velocity $u$ at an angle $\theta$ with the horizontal is given by the formula: $R = \frac{u^2 \sin(2\theta)}{g}$.
$1$. The value of the range $R$ depends on:
- The initial velocity $(u)$: $R$ is directly proportional to the square of the initial velocity $(u^2)$.
- The angle of projection $(\theta)$: $R$ depends on the sine of twice the angle of projection $(\sin(2\theta))$.
- The acceleration due to gravity $(g)$: $R$ is inversely proportional to the acceleration due to gravity $(g)$.
$2$. The maximum value of the range $(R_{max})$ depends on:
- The initial velocity $(u)$: $R_{max} = \frac{u^2}{g}$. Thus,it depends on the square of the initial velocity $(u^2)$.
- The acceleration due to gravity $(g)$: $R_{max}$ is inversely proportional to $g$.

(C) At the maximum height of a projectile,the vertical component of velocity $(v_y)$ becomes $0$,while the horizontal component $(v_x)$ remains constant $(u \cos \theta)$.
Thus,the resultant velocity vector is purely horizontal.
The acceleration of a projectile is always the acceleration due to gravity $(g)$,which acts vertically downwards.
Since the velocity is horizontal and the acceleration is vertical,the angle between them is $90^{\circ}$.

(C) Both balls will reach the ground at the same time.
Since both balls are released from the same height $h$ and their initial vertical velocity components are zero $(u_y = 0)$,the time taken to reach the ground is determined by the vertical motion equation: $h = \frac{1}{2}gt^2$.
Solving for $t$,we get $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are the same for both,the time $t$ is identical for both balls.

(B) $(i)$ $A$ projectile has its minimum speed at the highest point of its trajectory,where the vertical component of velocity becomes zero and only the horizontal component $u_x = u \cos \theta$ remains.
$(ii)$ The speed is maximum at the point of projection and at the point where the projectile hits the ground,as the vertical component of velocity is at its maximum magnitude at these points.

(B) For a projectile to achieve maximum range,the angle of projection must be $\theta = 45^{\circ}$.
At the maximum height,the vertical component of velocity becomes zero,and the projectile only possesses the horizontal component of velocity.
The horizontal component of velocity is given by $v_x = v_0 \cos \theta$.
Substituting $\theta = 45^{\circ}$,we get $v_x = v_0 \cos 45^{\circ}$.
Since $\cos 45^{\circ} = 1 / \sqrt{2}$,the velocity at maximum height is $v = v_0 / \sqrt{2}$.

(B) The horizontal range of a projectile is given by $R = \frac{v_0^2 \sin(2\theta_0)}{g}$.
The maximum height of a projectile is given by $H = \frac{v_0^2 \sin^2(\theta_0)}{2g}$.
Given that $R = nH$,we substitute the formulas:
$\frac{v_0^2 (2 \sin \theta_0 \cos \theta_0)}{g} = n \left( \frac{v_0^2 \sin^2 \theta_0}{2g} \right)$.
Canceling common terms $v_0^2$,$g$,and $\sin \theta_0$ from both sides:
$2 \cos \theta_0 = n \left( \frac{\sin \theta_0}{2} \right)$.
Rearranging to solve for $\tan \theta_0$:
$\frac{\sin \theta_0}{\cos \theta_0} = \frac{4}{n}$.
Therefore,$\tan \theta_0 = \frac{4}{n}$,which implies $\theta_0 = \tan^{-1}(\frac{4}{n})$.

(B) The maximum height $H$ attained by a projectile is given by $H = \frac{v_0^2 \sin^2 \theta}{2g}$.
For maximum height,$\theta = 90^\circ$,so $\sin \theta = 1$.
Thus,$H_{\text{max}} = \frac{v_0^2}{2g} = h$.
From this,we get $v_0^2 = 2gh$.
The horizontal range $R$ is given by $R = \frac{v_0^2 \sin(2\theta)}{g}$.
For maximum horizontal range,$\theta = 45^\circ$,so $\sin(2\theta) = \sin(90^\circ) = 1$.
Thus,$R_{\text{max}} = \frac{v_0^2}{g}$.
Substituting $v_0^2 = 2gh$ into the equation for $R_{\text{max}}$,we get $R_{\text{max}} = \frac{2gh}{g} = 2h$.

(A) The time of flight $T$ for a projectile is given by the formula $T = \frac{2 v_0 \sin \theta}{g}$.
For the first body projected at angle $\theta$,the time of flight is $T_1 = \frac{2 v_0 \sin \theta}{g}$.
For the second body projected at angle $(90^\circ - \theta)$,the time of flight is $T_2 = \frac{2 v_0 \sin(90^\circ - \theta)}{g} = \frac{2 v_0 \cos \theta}{g}$.
The ratio of their times of flight is $\frac{T_1}{T_2} = \frac{\frac{2 v_0 \sin \theta}{g}}{\frac{2 v_0 \cos \theta}{g}} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Thus,the ratio is $\tan \theta : 1$.

(B) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{v_{0}^{2} \sin(2\theta)}{g}$.
For the first case,$\theta_{1} = 15^{\circ}$ and $R_{1} = 1.5 \ km$.
$1.5 = \frac{v_{0}^{2} \sin(2 \times 15^{\circ})}{g} = \frac{v_{0}^{2} \sin(30^{\circ})}{g} = \frac{v_{0}^{2}}{2g}$.
Thus,$\frac{v_{0}^{2}}{g} = 1.5 \times 2 = 3.0 \ km$.
For the second case,$\theta_{2} = 45^{\circ}$.
The range $R_{2} = \frac{v_{0}^{2} \sin(2 \times 45^{\circ})}{g} = \frac{v_{0}^{2} \sin(90^{\circ})}{g} = \frac{v_{0}^{2}}{g}$.
Substituting the value of $\frac{v_{0}^{2}}{g}$ from the first case,we get $R_{2} = 3.0 \ km$.

(N/A) $v_{x} = \text{Horizontal component of velocity} = v \cos \theta = \text{constant}$
$v_{y} = \text{Vertical component of velocity} = v \sin \theta$
Velocity is always tangential to the curve in the direction of motion,and acceleration is always vertically downward and equal to $g$.

(N/A) The path of the ball as observed by a person standing on the footpath is parabolic.
When the ball is tossed,it possesses two components of velocity: a vertical component $(u_y)$ due to the toss and a horizontal component $(u_x)$ equal to the constant velocity of the car.
Since there is no horizontal acceleration acting on the ball,the horizontal velocity remains constant throughout the flight.
Simultaneously,the ball is under the influence of gravity,which causes a constant downward acceleration $(g)$.
The combination of uniform horizontal motion and uniformly accelerated vertical motion results in a parabolic trajectory for the ball relative to the stationary observer on the footpath.

(D) Let the velocity of the fielder be $\vec{u} = u\hat{i}$ and the velocity of the ball relative to the fielder be $\vec{v}_{rel} = v_{0}\cos\theta\hat{i} + v_{0}\sin\theta\hat{j}$.
$(a)$ The velocity of the ball relative to the ground is $\vec{v} = (u + v_{0}\cos\theta)\hat{i} + (v_{0}\sin\theta)\hat{j}$. The effective angle $\alpha$ is given by $\tan\alpha = \frac{v_{0}\sin\theta}{u + v_{0}\cos\theta}$,so $\alpha = \tan^{-1}\left(\frac{v_{0}\sin\theta}{u + v_{0}\cos\theta}\right)$.
$(b)$ The vertical component of velocity is $v_{y} = v_{0}\sin\theta$. The time of flight $T$ is given by $T = \frac{2v_{y}}{g} = \frac{2v_{0}\sin\theta}{g}$.
$(c)$ The horizontal range $R$ is $R = v_{x}T = (u + v_{0}\cos\theta)\left(\frac{2v_{0}\sin\theta}{g}\right) = \frac{2uv_{0}\sin\theta + v_{0}^{2}\sin(2\theta)}{g}$.
$(d)$ To maximize $R$,set $\frac{dR}{d\theta} = 0$: $\frac{d}{d\theta}\left[\frac{2uv_{0}\sin\theta + v_{0}^{2}\sin(2\theta)}{g}\right] = 0 \Rightarrow 2uv_{0}\cos\theta + 2v_{0}^{2}\cos(2\theta) = 0 \Rightarrow v_{0}\cos(2\theta) = -u\cos\theta$.
$(e)$ Solving $v_{0}(2\cos^{2}\theta - 1) = -u\cos\theta$,we get $2v_{0}\cos^{2}\theta + u\cos\theta - v_{0} = 0$. Using the quadratic formula for $\cos\theta$: $\cos\theta = \frac{-u + \sqrt{u^{2} + 8v_{0}^{2}}}{4v_{0}}$. As $u$ increases,$\cos\theta$ decreases,meaning $\theta$ increases.
$(f)$ For $u=0$,$\theta = 45^{\circ}$. If $u > 0$,$\theta < 45^{\circ}$ to compensate for the added horizontal velocity $u$.

(A) At the maximum height of a projectile's trajectory,the vertical component of velocity $(v_y)$ becomes zero. Thus,$(1)$ matches with $(b)$.
The linear velocity vector at any point on the trajectory is always directed tangent to the parabolic path. Thus,$(2)$ matches with $(a)$.
Therefore,the correct matching is $(1-b, 2-a)$.

(A) For a projectile launched horizontally,the initial velocity vector is parallel to the ground,so the angle of projection $\theta = 0^o$. Thus,$(1)$ matches with $(b)$.
In projectile motion,the only acceleration acting on the object is the acceleration due to gravity $(g)$,which acts vertically downwards. Therefore,the horizontal component of acceleration is $0$. Thus,$(2)$ matches with $(a)$.
The correct matching is $(1-b, 2-a)$.

(C) The formula for the maximum height $(h)$ attained by a projectile is given by:
$h = \frac{u^2 \sin^2 \theta}{2g}$
Since the angle of projection $(\theta)$ and acceleration due to gravity $(g)$ remain unchanged,the maximum height is directly proportional to the square of the initial velocity $(u)$:
$h \propto u^2$
Using the concept of relative error or percentage change for a power function:
$\frac{\Delta h}{h} \times 100 = 2 \times \left( \frac{\Delta u}{u} \times 100 \right)$
Given that the percentage change in velocity is $\frac{\Delta u}{u} \times 100 = 2 \%$,we substitute this into the equation:
$\frac{\Delta h}{h} \times 100 = 2 \times 2 \% = 4 \%$
Therefore,the percentage change in the maximum height attained is $4 \%$.

(A) The given equation of the trajectory is $y = \alpha x - \beta x^2$.
Comparing this with the standard equation of a projectile trajectory: $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
By comparing the coefficients of $x$,we get $\tan \theta = \alpha$,which implies $\theta = \tan^{-1} \alpha$.
By comparing the coefficients of $x^2$,we get $\beta = \frac{g}{2u^2 \cos^2 \theta}$.
Rearranging for $u^2$,we find $u^2 = \frac{g}{2\beta \cos^2 \theta}$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting $u^2$ into the formula: $H = \frac{g}{2\beta \cos^2 \theta} \cdot \frac{\sin^2 \theta}{2g} = \frac{\tan^2 \theta}{4\beta}$.
Since $\tan \theta = \alpha$,we have $H = \frac{\alpha^2}{4\beta}$.

(C) The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the values: $H = \frac{(25)^2 \cdot (\sin 45^{\circ})^2}{2 \times 10} = \frac{625 \times 0.5}{20} = \frac{312.5}{20} = 15.625\, m$.
The formula for time taken to reach the highest point is $T = \frac{u \sin \theta}{g}$.
Substituting the values: $T = \frac{25 \times \sin 45^{\circ}}{10} = \frac{25 \times 0.707}{10} = 2.5 \times 0.707 = 1.7675\, s \approx 1.77\, s$.