Projectile Motion from Hight Questions in English

(A) Given: Initial velocity $u = 10 \, m/s$,angle of projection $\theta = 30^{\circ}$,and vertical displacement $h = -60 \, m$ (downward).
Using the equation of motion in the vertical direction: $s_y = u_y T + \frac{1}{2} a_y T^2$
Here,$u_y = u \sin \theta = 10 \sin 30^{\circ} = 10 \times 0.5 = 5 \, m/s$,$a_y = -g = -10 \, m/s^2$,and $s_y = -60 \, m$.
Substituting the values: $-60 = 5T - \frac{1}{2} \times 10 \times T^2$
$-60 = 5T - 5T^2$
Dividing by $-5$: $T^2 - T - 12 = 0$
Factoring the quadratic equation: $(T - 4)(T + 3) = 0$
Since time $T$ cannot be negative,$T = 4 \, s$.

(B) The initial horizontal velocity is $u_x = 5 \ m/s$ and the initial vertical velocity is $u_y = 0 \ m/s$.
The vertical velocity $v_y$ when the stone hits the ground is given by the equation $v_y^2 = u_y^2 + 2gh$.
Substituting the values,$v_y^2 = 0^2 + 2 \times 10 \times 10 = 200$,so $v_y = \sqrt{200} \ m/s$.
The horizontal velocity remains constant at $v_x = 5 \ m/s$.
The net speed $v$ when hitting the ground is $v = \sqrt{v_x^2 + v_y^2}$.
$v = \sqrt{5^2 + 200} = \sqrt{25 + 200} = \sqrt{225} = 15 \ m/s$.

(D) Let the angles with the horizontal be $\theta_A$ and $\theta_B$. Given angles with vertical are $45^{\circ}$ and $60^{\circ}$,so $\theta_A = 90^{\circ} - 45^{\circ} = 45^{\circ}$ and $\theta_B = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
For a projectile thrown from height $h$,the time of flight $T$ is given by $h = -u \sin \theta T + \frac{1}{2} g T^2$. Since $T$ is the same for both,the vertical components of initial velocity must be equal: $v_A \sin \theta_A = v_B \sin \theta_B$.
$v_A \sin 45^{\circ} = v_B \sin 30^{\circ} \implies v_A (\frac{1}{\sqrt{2}}) = v_B (\frac{1}{2}) \implies \frac{v_A}{v_B} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Also,the range $R = (v \cos \theta) T$. Since $R$ and $T$ are same,$v_A \cos \theta_A = v_B \cos \theta_B$.
$v_A \cos 45^{\circ} = v_B \cos 30^{\circ} \implies v_A (\frac{1}{\sqrt{2}}) = v_B (\frac{\sqrt{3}}{2}) \implies \frac{v_A}{v_B} = \frac{\sqrt{6}}{2} = \sqrt{\frac{3}{2}}$.
Since the two conditions lead to different ratios,the problem statement is physically inconsistent. However,if we assume the vertical components are equal to satisfy the time of flight condition,the ratio is $1 : \sqrt{2}$.

(D) Given:
Horizontal speed of helicopter,$u = 360 \ km/h = 360 \times \frac{5}{18} = 100 \ m/s$.
Altitude,$H = 2 \ km = 2000 \ m$.
Time taken to hit the ground,$t = 20 \ s$.
Step $1$: Calculate the horizontal displacement $(x)$ of the object.
$x = u \times t = 100 \ m/s \times 20 \ s = 2000 \ m = 2 \ km$.
Step $2$: The vertical displacement is the altitude $H = 2 \ km$.
Step $3$: Calculate the total displacement $(D)$ from the release point to point $O$.
$D = \sqrt{x^2 + H^2} = \sqrt{(2 \ km)^2 + (2 \ km)^2} = \sqrt{4 + 4} \ km = \sqrt{8} \ km = 2\sqrt{2} \ km$.
Thus,the correct option is $D$.

(A) The horizontal component of the velocity of the bomb remains constant throughout the motion because there is no acceleration in the horizontal direction.
$V_x = u = 500 \ m/s$
The vertical component of the velocity at time $t = 10 \ s$ can be calculated using the first equation of motion:
$V_y = u_y + g t$
Since the bomb is released horizontally,the initial vertical velocity $u_y = 0$.
$V_y = 0 + (10 \ m/s^2) \times (10 \ s) = 100 \ m/s$
The angle $\theta$ that the velocity vector makes with the horizontal is given by:
$\tan \theta = \frac{V_y}{V_x}$
$\tan \theta = \frac{100}{500} = \frac{1}{5}$
$\theta = \tan^{-1}\left(\frac{1}{5}\right)$

(B) The initial vertical velocity of the body is $u_y = 0 \ m/s$. The height is $h = 1960 \ m$. Using the equation of motion $h = \frac{1}{2} gt^2$,we can find the time $t$ taken to reach the ground:
$1960 = \frac{1}{2} \times 9.8 \times t^2$
$t^2 = \frac{1960 \times 2}{9.8} = 400$
$t = 20 \ s$
Now,convert the horizontal velocity $v$ from $km/h$ to $m/s$:
$v = 540 \times \frac{5}{18} = 150 \ m/s$
The horizontal distance $AB$ covered by the body is given by $AB = v \times t$:
$AB = 150 \times 20 = 3000 \ m$

(D) The initial vertical component of velocity is $u_y = u \sin \theta = 15 \sin 30^{\circ} = 15 \times 0.5 = 7.5 \,ms^{-1}$.
Using the equation of motion for vertical displacement $y = u_y t + \frac{1}{2} a_y t^2$, where $y = -h$ (downward displacement), $u_y = 7.5 \,ms^{-1}$, $a_y = -g = -10 \,ms^{-2}$, and $t = 3 \,s$:
$-h = (7.5)(3) + \frac{1}{2}(-10)(3)^2$
$-h = 22.5 - 5(9)$
$-h = 22.5 - 45$
$-h = -22.5$
$h = 22.5 \,m$.
Therefore, the height of the building is $22.5 \,m$.

(B) According to the law of conservation of mechanical energy,the total mechanical energy at the top of the cliff is equal to the total mechanical energy at the ground level.
Initial kinetic energy + Initial potential energy = Final kinetic energy at ground
$\frac{1}{2} m v_1^2 + m g h = \frac{1}{2} m v_2^2$
Dividing both sides by $\frac{1}{2} m$,we get:
$v_1^2 + 2 g h = v_2^2$
$v_2 = \sqrt{v_1^2 + 2 g h}$
Given:
Initial speed $v_1 = 50 \ m \ s^{-1}$
Height $h = 55 \ m$
Acceleration due to gravity $g = 10 \ m \ s^{-2}$
Substituting the values:
$v_2 = \sqrt{(50)^2 + 2 \times 10 \times 55}$
$v_2 = \sqrt{2500 + 1100}$
$v_2 = \sqrt{3600}$
$v_2 = 60 \ m \ s^{-1}$

(D) For a projectile launched horizontally from a height $h$ with initial velocity $u = 10 \sqrt{3} \text{ m s}^{-1}$:
$1$. The horizontal component of velocity remains constant: $v_x = u = 10 \sqrt{3} \text{ m s}^{-1}$.
$2$. The vertical component of velocity at the time of landing is $v_y = \sqrt{2gh}$.
$3$. The angle $\theta$ that the velocity vector makes with the horizontal is given by $\tan \theta = \frac{v_y}{v_x}$.
$4$. Given that the landing angle is $30^{\circ}$ or more,we have $\tan \theta \geq \tan 30^{\circ}$.
$5$. Substituting the values: $\frac{\sqrt{2gh}}{10 \sqrt{3}} \geq \frac{1}{\sqrt{3}}$.
$6$. Simplifying: $\sqrt{2gh} \geq 10$.
$7$. Squaring both sides: $2gh \geq 100$.
$8$. With $g = 10 \text{ m s}^{-2}$,we get $20h \geq 100$,which implies $h \geq 5 \text{ m}$.
$9$. The minimum height is $5 \text{ m}$.

(B) We consider the vertical motion of the ball.
Initial vertical velocity: $u_y = u \sin \theta = 20 \times \sin 30^{\circ} = 20 \times \frac{1}{2} = 10 \,m/s$.
Vertical acceleration: $a_y = -g = -10 \,m/s^2$.
Time taken to reach the ground: $t = 3 \,s$.
Using the equation of motion for vertical displacement: $s_y = u_y t + \frac{1}{2} a_y t^2$.
Here, the displacement $s_y = -h$ (where $h$ is the height of the building).
$-h = (10)(3) + \frac{1}{2}(-10)(3)^2$.
$-h = 30 - 5(9) = 30 - 45 = -15 \,m$.
Therefore, $h = 15 \,m$.

(A) The horizontal component of velocity remains constant because there is no acceleration in the horizontal direction: $v_x = u_x = 15 \ m \ s^{-1}$.
In the vertical direction,the stone undergoes free fall with initial vertical velocity $u_y = 0$. Using the equation $v_y^2 = u_y^2 + 2gh$:
$v_y^2 = 0 + 2 \times 9.8 \times 490 = 9604$.
$v_y = \sqrt{9604} = 98 \ m \ s^{-1}$.
The final speed $v$ is the magnitude of the resultant velocity vector: $v = \sqrt{v_x^2 + v_y^2}$.
$v = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx 99.14 \ m \ s^{-1}$.
Rounding to the nearest integer,the speed is $99 \ m \ s^{-1}$.

(C) Given: Initial horizontal velocity $u_x = 20 \ ms^{-1}$,initial vertical velocity $u_y = 0 \ ms^{-1}$,acceleration $g = 10 \ ms^{-2}$,time $t = 1 \ s$.
$A$. Velocity of the body after $1 \ s$:
$v_x = u_x = 20 \ ms^{-1}$
$v_y = u_y + gt = 0 + 10(1) = 10 \ ms^{-1}$
Resultant velocity $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 10^2} = \sqrt{500} \approx 22.4 \ ms^{-1}$. Thus,$A-IV$.
$B$. Horizontal displacement after $1 \ s$:
$x = u_x t = 20 \times 1 = 20 \ m$. Thus,$B-II$.
$C$. Vertical displacement after $1 \ s$:
$y = u_y t + \frac{1}{2}gt^2 = 0 + \frac{1}{2}(10)(1)^2 = 5 \ m$. Thus,$C-I$.
$D$. Vertical velocity after $1 \ s$:
$v_y = u_y + gt = 0 + 10(1) = 10 \ ms^{-1}$. Thus,$D-III$.
Therefore,the correct matching is $A-IV, B-II, C-I, D-III$.

(A) Given, initial velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ m/s}$.
Height of the tower $h = 30 \text{ m}$. Since $\hat{k}$ is vertically upward, the vertical displacement $S_z = -30 \text{ m}$ and acceleration $a_z = -g = -10 \text{ m/s}^2$.
The initial vertical velocity component is $u_z = 5 \text{ m/s}$.
Using the equation of motion $S_z = u_z t + \frac{1}{2} a_z t^2$:
$-30 = 5t - \frac{1}{2} \times 10 \times t^2$
$-30 = 5t - 5t^2$
$t^2 - t - 6 = 0$
$(t - 3)(t + 2) = 0$
Since time cannot be negative, $t = 3 \text{ s}$.
In the horizontal plane, the velocity components are $v_x = 3 \text{ m/s}$ (East) and $v_y = 4 \text{ m/s}$ (North).
The horizontal displacements in $3 \text{ s}$ are:
$x = v_x \times t = 3 \times 3 = 9 \text{ m}$
$y = v_y \times t = 4 \times 3 = 12 \text{ m}$
The horizontal range $R$ is the distance from the base of the tower:
$R = \sqrt{x^2 + y^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \text{ m}$.

(A) The initial velocity $u = 100 \ ms^{-1}$ at an angle $\theta = 30^{\circ}$ above the horizontal.
The vertical component of velocity is $u_y = u \sin 30^{\circ} = 100 \times 0.5 = 50 \ ms^{-1}$.
The horizontal component of velocity is $u_x = u \cos 30^{\circ} = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \ ms^{-1}$.
Taking the downward direction as positive,the vertical displacement $h = 375 \ m$ and acceleration $g = 10 \ ms^{-2}$.
Using the equation of motion $h = u_y t + \frac{1}{2} g t^2$:
$375 = -50t + \frac{1}{2} \times 10 \times t^2$
$375 = -50t + 5t^2$
$5t^2 - 50t - 375 = 0$
Dividing by $5$: $t^2 - 10t - 75 = 0$
$(t - 15)(t + 5) = 0$
Since time cannot be negative,$t = 15 \ s$.
The horizontal distance $x = u_x \times t = 50\sqrt{3} \times 15 = 750\sqrt{3} \ m$.

(B) Since the hill is smooth,the potential energy at the top is converted into kinetic energy at the point where the height is $\frac{H}{2}$.
Let $v$ be the velocity at height $\frac{H}{2}$. By the law of conservation of energy:
$mgH = mg(\frac{H}{2}) + \frac{1}{2}mv^2$
$mg(\frac{H}{2}) = \frac{1}{2}mv^2$
$v = \sqrt{gH}$
Now,the object acts as a horizontal projectile from a height $h' = \frac{H}{2}$.
The time taken to reach the ground is given by:
$t = \sqrt{\frac{2h'}{g}} = \sqrt{\frac{2(H/2)}{g}} = \sqrt{\frac{H}{g}}$
The horizontal distance (range) covered is:
$x = v \times t = \sqrt{gH} \times \sqrt{\frac{H}{g}} = H$

(B) Let the initial position be at height $h = 4.2 \text{ m}$. The initial velocity components are $u_x = v \cos 30^{\circ} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \text{ m/s}$ and $u_y = v \sin 30^{\circ} = 40 \times \frac{1}{2} = 20 \text{ m/s}$.
Using the equation of motion for the vertical direction,taking the downward direction as positive:
$y = u_y t + \frac{1}{2} g t^2$
$4.2 = -20t + \frac{1}{2} (10) t^2$
$5t^2 - 20t - 4.2 = 0$
Multiplying by $10$ to simplify: $50t^2 - 200t - 42 = 0$. This is incorrect; let's use the standard convention where upward is positive:
$-4.2 = 20t - \frac{1}{2} (10) t^2$
$5t^2 - 20t - 4.2 = 0$
$t = \frac{20 \pm \sqrt{(-20)^2 - 4(5)(-4.2)}}{2(5)} = \frac{20 \pm \sqrt{400 + 84}}{10} = \frac{20 \pm \sqrt{484}}{10} = \frac{20 \pm 22}{10}$.
Since $t > 0$,$t = \frac{42}{10} = 4.2 \text{ s}$.
The horizontal distance $R$ is given by $R = u_x t = (20\sqrt{3}) \times 4.2 = 84\sqrt{3} \text{ m}$.

(B) Given: Initial velocity $U_1 = U_2 = 2 \sqrt{2} \ m \ s^{-1}$,acceleration due to gravity $g = 10 \ m \ s^{-2}$.
At time $t$,the velocity vectors are given by:
$\vec{v}_1 = (U_1 \cos 45^{\circ}) \hat{i} + (U_1 \sin 45^{\circ} - gt) \hat{j} = (2 \sqrt{2} \cdot \frac{1}{\sqrt{2}}) \hat{i} + (2 \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 10t) \hat{j} = 2 \hat{i} + (2 - 10t) \hat{j} \dots(1)$
$\vec{v}_2 = (U_2 \cos 135^{\circ}) \hat{i} + (U_2 \sin 135^{\circ} - gt) \hat{j} = (2 \sqrt{2} \cdot -\frac{1}{\sqrt{2}}) \hat{i} + (2 \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 10t) \hat{j} = -2 \hat{i} + (2 - 10t) \hat{j} \dots(2)$
Since the velocity vectors are perpendicular,their dot product must be zero:
$\vec{v}_1 \cdot \vec{v}_2 = 0$
$(2 \hat{i} + (2 - 10t) \hat{j}) \cdot (-2 \hat{i} + (2 - 10t) \hat{j}) = 0$
$-4 + (2 - 10t)^2 = 0$
$(2 - 10t)^2 = 4$
Taking the square root on both sides:
$2 - 10t = \pm 2$
Case $1$: $2 - 10t = 2 \Rightarrow 10t = 0 \Rightarrow t = 0 \ s$
Case $2$: $2 - 10t = -2 \Rightarrow 10t = 4 \Rightarrow t = 0.4 \ s$
Thus,the velocity vectors are perpendicular at $t = 0.4 \ s$.

(C) The force acting on the ball is gravity,which acts only in the vertical $(y)$ direction. Therefore,there is no change in momentum in the horizontal $(x)$ direction.
Initial vertical velocity: $u_y = u \sin 45^{\circ} = \sqrt{10} \cdot \frac{1}{\sqrt{2}} = \sqrt{5} \ m/s$.
Initial vertical momentum: $p_{yi} = m u_y = 0.2 \cdot \sqrt{5} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}} \ kg \cdot m/s$ (upwards).
Using the equation of motion $v_y^2 = u_y^2 + 2gh$ (taking downward as positive for final velocity calculation,or using energy conservation):
$v_y^2 = (\sqrt{5})^2 + 2(10)(1) = 5 + 20 = 25$.
So,final vertical velocity $v_y = 5 \ m/s$ (downwards).
Final vertical momentum: $p_{yf} = m v_y = 0.2 \cdot 5 = 1 \ kg \cdot m/s$ (downwards).
Taking upward direction as positive,$p_{yi} = \frac{1}{\sqrt{5}}$ and $p_{yf} = -1$.
The change in momentum is $\Delta p = p_{yf} - p_{yi} = -1 - \frac{1}{\sqrt{5}} = -(1 + \frac{1}{\sqrt{5}})$.
The modulus of the momentum increment is $|\Delta p| = 1 + \frac{1}{\sqrt{5}} = \frac{\sqrt{5}+1}{\sqrt{5}}$.

(B) The change in potential energy $(\Delta PE)$ is given by $\Delta PE = mgh$, where $h$ is the vertical displacement.
To find the vertical displacement, we consider the vertical component of the motion:
Initial vertical velocity, $u_y = 24 \, m/s$
Acceleration, $a_y = -g = -10 \, m/s^2$
Time, $t = 6 \, s$
Using the equation of motion $h = u_y t + \frac{1}{2} a_y t^2$:
$h = (24)(6) + \frac{1}{2}(-10)(6)^2$
$h = 144 - 5(36)$
$h = 144 - 180 = -36 \, m$
Now, calculate the change in potential energy:
$\Delta PE = mgh = (1 \, kg)(10 \, m/s^2)(-36 \, m)$
$\Delta PE = -360 \, J$