Question

A man is standing on a rail car travelling with a constant speed of v = 10 m/s. He wishes to throw a ball through a stationary hoop 5 m above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of 12.5 m/s w.r.t himself.

(a) What must be the vertical component of the initial velocity of the ball? ($v_{perp}$)

(b) How many seconds after he releases the ball will it pass through the hoop? ($T$)

(c) At what horizontal distance in front of the loop must he release the ball? ($s_x$)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The important aspects to be noticed in this problem are:

(1) Velocity of projection of ball is relative to man in motion.

(2) Ball clears the hoop when it is at the topmost point.

$\overrightarrow{v_{ball/man}}$ = $\overrightarrow{v_{ball}}$ - $\overrightarrow{v_{man}}\overrightarrow{v_{ball}}$ = $\overrightarrow{v_{ball/man}}$ + $\overrightarrow{v_{man}}$

(a) Now we apply the above relation to x as well as y-component of velocity. If ball is projected with velocity $v_0$ and angle $\theta$, then

x-component of $\overrightarrow{v_{ball}}$ = $(v_{0}cos\theta +10)\frac{m}{s}$ y-component of $\overrightarrow{v_{ball}}$ = $\left(v_{0}sin\theta \right)\frac{m}{s}$

(b) Since vertical component of ball's velocity is unaffected by horizontal motion of car, we can use the formula for time of flight.

$\frac{(12.5sin\theta {)}^2}{2g}$ = $5msi{n}^2\theta$ = $\frac{5\times (2\times 10)}{12.5\times 12.5}sin\theta$ = $\frac{4}{5}$ and $cos\theta$ = $\frac{3}{5}$; $v_{0}sin\theta$ = $\left(12.5\right)\times \left(\frac{4}{5}\right)$

= $10\frac{m}{s}$

Time taken to reach maximum height = $\frac{sin\theta }{g}$ = $\frac{10}{10}$ = $1sec$

(c) Horizontal distance of loop from point of projection = $(12.5cos\theta +10)\times 1$ = $17.5m$