Question

A projectile initially has the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3m{s^{ - 2}}$ for $0.5\min$ .If the maximum height reached by it is $80m$ , then the angle of projection is ( $g = 10m{s^{ - 2}}$ )
(A) ${\tan ^{ - 1}}3$
(B) ${\tan ^{ - 1}}(\dfrac{3}{2})$
(C) ${\tan ^{ - 1}}(\dfrac{4}{9})$
(D) ${\sin ^{ - 1}}(\dfrac{4}{9})$

Answer 1

Hint : Find the horizontal velocity of the particle using Equations for 1D motion and use the equation for maximum height reached by a projectile to find the angle of projection. The maximum height of a projectile is given by, $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ .where is the $u$ initial velocity $\theta$ is the angle of projection, $g$ is the gravitational acceleration and $u\sin \theta$ is the initial velocity along the vertical.

Complete Step By Step Answer:
We know that the maximum height of a projectile is given by, $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ where is the $u$ initial velocity $\theta$ is the angle of projection, $g$ is the gravitational acceleration and $u\sin \theta$ is the initial velocity along the vertical. Here , we have given, the maximum height reached by the projectile is, $H = 80m$ . And gravitational acceleration on the body is, $g = 10m{s^{ - 2}}$ .
putting the values in the equation we get, $80 = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times 10}}$
Or, ${u^2}{\sin ^2}\theta = 1600$
Or, $u\sin \theta = 40m{s^{ - 1}}$
Now, we have to find the horizontal component of the velocity to find the angle of the projection.
Here, we have given , the projectile would have the same velocity if it was accelerated at a rate of for $0.5\min = 30\sec$ .Therefore, using equation of rectilinear motion $v = u + at$ we get,
here, $u = 0$ , $v = u\cos \theta$ , $a = 3$ and $t = 30$ . Putting these values we get,
$u\cos \theta = 3 \times 30 = 90m{s^{ - 1}}$
Therefore, the ratio of these to velocity components is,
$\dfrac{{u\sin \theta }}{{u\cos \theta }} = \dfrac{{40}}{{90}}$
Hence, $\tan \theta = \dfrac{4}{9}$
Therefore, angle of projection is, $\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{9}} \right)$
Hence, option (C ) is correct.

Note :
$\bullet$ The total time taken by a body moving in a straight line with same velocity as the horizontal velocity of a projectile is the same also the displacement of the body is equal to the range of the projectile. $\bullet$ For a projectile released from a height $H$ , the maximum height reached by the projectile with respect to height is the same for as the projectile launched from the ground with the same angle of projection the time of flight is differ due to the extra fall from height $H$ .