Question

Two identical balls are set into motion simultaneously from equal heights h. While the ball $A$ is thrown horizontally with velocity $v$, the ball $B$ is just released to fall by itself. Choose the alternative that best represents the motion of $A$ and $B$ with respect to an observer who moves with velocity $\frac{v}{2}$ with respect to the ground as shown in the figure.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$A$ is basically a projectile

$v_x$ = $v\hat{i}$

$v_y$ = $v_{y}t+\frac{1}{2}{a}_y{t}^2$ $\because v_y$ = $0$

$a_y$ = $-g$

$\Rightarrow v_y$ = $-\frac{1}{2}g{t}^2\hat{j}$

$\Rightarrow v_A$ = $v\hat{i}-\frac{1}{2}g{t}^2\hat{j}$

$B$ is a ball falling down $u_x$ = $0$ $v_y$ = $0$ $a_x$ = $0$ $a_y$ = $-g$

No motion in $x$ direction

In $y$ direction

$v_y$ = $-\frac{1}{2}g{t}^2\hat{j}$

$\Rightarrow v_B$ = $-\frac{1}{2}g{t}^2\hat{j}$

Man's velocity

Man is moving with velocity $\frac{v}{2}$ along the x-axis

$\Rightarrow v_M$ = $+\frac{v}{2}\hat{i}$

Now we have to find $v_{\frac{A}{M}}$ & $v_{\frac{B}{M}}$

$\Rightarrow$ $v_{\frac{A}{M}}$ = $v_A-v_M$

= $v\hat{i}-\frac{1}{2}g{t}^2\hat{j}-\frac{v}{2}\hat{i}$

= $\frac{v}{2}\hat{i}-\frac{1}{2}g{t}^2\hat{j}$
$v_{B/M}=v_B-V_M$
$v_{B/M}=-\frac{1}{2}g{t}^2\hat{j}-\frac{v}{2}\hat{i}$
so option "c" is correct