(B) The initial velocity components are $u_x = 20 \cos \alpha$ and $u_y = 20 \sin \alpha$.Since the horizontal acceleration is zero,the horizontal vel

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(B) The initial velocity components are $u_x = 20 \cos \alpha$ and $u_y = 20 \sin \alpha$.Since the horizontal acceleration is zero,the horizontal velocity remains constant: $v_x = u_x = 20 \cos \alpha$.The vertical velocity after time $t = 10 \; s$ is given by $v_y = u_y - gt = 20 \sin \alpha - 10 \times 10 = 20 \sin \alpha - 100$.The inclination $\beta$ with the horizontal is given by $\tan \beta = \frac{v_y}{v_x}$.Substituting the values: $\tan \beta = \frac{20 \sin \alpha - 100}{20 \cos \alpha} = \frac{20 \sin \alpha}{20 \cos \alpha} - \frac{100}{20 \cos \alpha} = \tan \alpha - 5 \sec \alpha$.

Class 11 Physics 3-2.Motion in Plane Mix Examples-Motion in Plane

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$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

JEE MAIN 2022 All JEE MAIN

A
$\tan \alpha + 5 \sec \alpha$
B
$\tan \alpha - 5 \sec \alpha$
C
$2 \tan \alpha - 5 \sec \alpha$
D
$2 \tan \alpha + 5 \sec \alpha$

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Mix Examples-Motion in Plane

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$A$ projectile is thrown into air with velocity $15 \ m/s$ at an angle $30^{\circ}$ with the horizontal. After what time is its direction of motion perpendicular to its initial direction (in $s$)? (Assume $g = 10 \ m/s^2$)

DifficultTS EAMCET 2020

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$A$ particle has an initial velocity of $(\hat{i} + \hat{j}) \, m/s$ and an acceleration of $(\hat{i} + \hat{j}) \, m/s^2$. Its speed after $10 \, s$ will be:

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$A$ ball of mass $1 \; kg$ is thrown vertically upwards and returns to the ground after $3 \; s$. Another ball,thrown at $60^{\circ}$ with the vertical,also stays in the air for the same time before it touches the ground. The ratio of the two maximum heights reached is:

MediumNEET 2017

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The trajectory of a particle in projectile motion is given by $y = x - \frac{x^2}{80}$. Here, $x$ and $y$ are in meters. For this projectile motion, match the following with $g = 10 \, m/s^2$.
$Column-I$ $Column-II$
$(A)$ Angle of projection $(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$ $(q)$ $80 \, m$
$(C)$ Maximum height $(r)$ $45^{\circ}$
$(D)$ Horizontal range $(s)$ $\tan^{-1}(1/2)$

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Assertion $(A)$: When a vehicle takes a turn on the road,it travels along a curved path. Reason $(R)$: In a curved path,the velocity of the vehicle remains the same.

EasyAP EAMCET 2017

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$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

Similar Questions

$A$ projectile is thrown into air with velocity $15 \ m/s$ at an angle $30^{\circ}$ with the horizontal. After what time is its direction of motion perpendicular to its initial direction (in $s$)? (Assume $g = 10 \ m/s^2$)

$A$ particle has an initial velocity of $(\hat{i} + \hat{j}) \, m/s$ and an acceleration of $(\hat{i} + \hat{j}) \, m/s^2$. Its speed after $10 \, s$ will be:

$A$ ball of mass $1 \; kg$ is thrown vertically upwards and returns to the ground after $3 \; s$. Another ball,thrown at $60^{\circ}$ with the vertical,also stays in the air for the same time before it touches the ground. The ratio of the two maximum heights reached is:

Assertion $(A)$: When a vehicle takes a turn on the road,it travels along a curved path. Reason $(R)$: In a curved path,the velocity of the vehicle remains the same.

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