Question

A particle projected from ground moves at angle $45^\circ$ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is (Neglect the effect of air resistance)
(A) ${\tan ^{ - 1}}\left( 3 \right)$
(B) ${\tan ^{ - 1}}\left( 2 \right)$
(C) ${\tan ^{ - 1}}\left( {\sqrt 2 } \right)$
(D) ${\tan ^{ - 1}}\left( 4 \right)$

Answer 1

At any point, the tan of the angle the particle makes with the vertical is the ratio of the vertical velocity to the horizontal velocity. The point of minimum speed is the maximum height.
Formula used: In this solution we will be using the following formulae;
$T = \dfrac{{u\sin \theta }}{g}$ where $T$ is the time taken to reach maximum height, $u$ is the initial velocity of projection, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.
$\tan \alpha = \dfrac{{{v_v}}}{{{v_h}}}$ where $\alpha$ is the angle the velocity of the particle makes with the horizontal at any point, ${v_v}$ is the vertical component of the velocity at the point, and ${v_h}$ is the horizontal component of the velocity.
${v_v} = u\sin \theta - gt$ where $t$ is the time taken to get to the point where it has the velocity ${v_v}$. ${v_h} = u\cos \theta$

Complete Step-by-Step Solution:
The point at which the particle has the minimum speed (which is actually zero for uninterrupted flight or journey) is the point of maximum height. Hence, according to the question it took 2 seconds to reach maximum height. Then, using the formula for time taken to reach maximum height, we have
$T = \dfrac{{u\sin \theta }}{g}$
From the question, we have
$2 = \dfrac{{u\sin \theta }}{g}$
$\Rightarrow u\sin \theta = 2g$
Now, at any point, the angle to the horizontal can be given as
$\tan \alpha = \dfrac{{{v_v}}}{{{v_h}}}$, where ${v_v}$ is the vertical component of the velocity at the point, and ${v_h}$ is the horizontal component of the velocity.
Hence at time 1 second, we can write
$\tan 45^\circ = \dfrac{{{v_v}}}{{{v_h}}}$
But ${v_v} = u\sin \theta - gt$, and ${v_h} = u\cos \theta$, then
$\tan 45^\circ = \dfrac{{u\sin \theta - gt}}{{u\cos \theta }}$
$\Rightarrow u\cos \theta = u\sin \theta - gt$ (since $\tan 45^\circ = 1$)
From above $u\sin \theta = 2g$,
Hence,
$u\cos \theta = 2g - gt$
But $t = 1$s. then,
$u\cos \theta = 2g - g = g$
Hence, dividing $u\sin \theta$ by $u\cos \theta$ we get
$\dfrac{{u\sin \theta }}{{u\cos \theta }} = \dfrac{{2g}}{g}$
$\Rightarrow \tan \theta = 2$
Finding the trigonometric inverse, we have
$\theta = {\tan ^{ - 1}}\left( 2 \right)$

Hence, the correct option is B

Note: For clarity, we see that the horizontal velocity is ${v_h} = u\cos \theta$ because whenever we neglect air resistance, the horizontal velocity is always constant, equal to the horizontal velocity of projection, which of course is given as $u\cos \theta$. But the vertical component reduces linearly with time, since it is against gravity. Hence, we have $u\sin \theta - gt$.