Question

A man is standing on a rail road car travelling with a constant speed of $v=10ms$${}^{-1}$ .He wishes to throw a ball through a stationary hoop 5 m above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of $12.5ms$${}^{-1}$ w.r.t. himself.
a. What must be the vertical component of the initial velocity of the ball?
b. How many seconds after he releases the ball will it pass through the hoop?
c. At what horizontal distance in front of the loop must he release the ball?

Answer 1

The important aspects to be noticed in this problem are:

The velocity of the projection of ball is relative to man in motion.

The 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 the ball is projected with velocity $v_0$ and angle $\theta$ then

x-component of ${\overrightarrow{v}}_{ball}=(v_{0}cos\theta +10)m{s}^{-1}$

y-component of ${\overrightarrow{v}}_{ball}=\left(v_{0}sin\theta \right)m{s}^{-1}$

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

$\frac{\left(12.5\sin \theta \right)}{2g}=5m$

${\sin }^2\theta =\frac{5\times (2\times 10)}{12.5\times 12.5}$

$\sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}$

$v_0\sin \theta =\left(12.5\right)\times \left(\frac{4}{5}\right)=10m{s}^{-1}$

Time taken to reach the maximum height,

$\frac{2v_0\sin \theta }{g}=\frac{2\times 10}{10}=2s$

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