Question

A ball is projected from the ground with a speed $u$ making an angle $\theta$ with the horizontal. Which of the following graphs correctly represents the variation of angle between the velocity and acceleration vectors with time.
[If $T$ is the time of flight]

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Velocity of the ball at any instant during the projectile motion is given by
$\overrightarrow{v}=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}$ and
$\overrightarrow{a}=-g\hat{j}$
Let $\varphi$ be the angle between $\overrightarrow{v}$ and $\overrightarrow{a}$.
$\cos \varphi =\frac{\overrightarrow{v}.\overrightarrow{a}}{\mid \overrightarrow{v}\mid \mid \overrightarrow{a}\mid }=\frac{g(gt-u\sin \theta )}{g\sqrt{u^2{\cos }^2\theta +u^2{\sin }^2\theta +g^2{t}^2-2ugt\sin \theta }}$
$\varphi ={\cos }^{-1}\left(\frac{gt-u\sin \theta }{\sqrt{u^2+g^2{t}^2-2ugt\sin \theta }}\right)$
Now, let us find the values of $\varphi$ at $t=0,\frac{T}{2}\text{and}T$
Time of flight, $T=\frac{2u\sin \theta }{g}$
At $t=0$, $\varphi =\frac{\pi }{2}+\theta$
At $t=\frac{T}{2},\varphi =\frac{\pi }{2}$
At $t=T,\varphi =\frac{\pi }{2}-\theta$
During ascent, the angle between velocity and acceleration is obtuse, then ${90}^{\circ }$ at the highest point and then acute during its desent.
Thus, option (B) is the correct answer