Question

A tennis ball launcher is placed on the floor at the front end of a train that has a uniform forward acceleration of 2.00 $m/{{s}^{2}}$. The launcher projects a ball at an initial speed of 20 m/s with respect to the train toward its rear. The ball achieves a maximum height of 10.0 m. Ignore air resistance. Find out how far in m, from the front end of the train the ball lands inside the train.
A. 24m
B. 48 m
C. 45 m
D. 96 m

Answer 1

The launcher is on the train and the train is moving with constant acceleration, so the launcher is also moving with constant acceleration. So, it has some initial velocity. The ball is projected towards the horizontal, so there is some inclination with the horizontal. The maximum height achieved is given. We can use the concept of projectile here.

Complete step by step answer:
The launcher is placed at the front and the ball is projected towards the back. The projectile in this case has 2 accelerations: one in downward direction, acceleration due to gravity and the other is in the horizontal, both are constant.
First of all, let us write all the parameters that we have:
${{u}_{x}}=20m/s$, ${{u}_{y}}=0$, ${{a}_{x}}=2m/{{s}^{2}},{{a}_{y}}=g=10m/{{s}^{2}}$
Maximum height achieved is 10m,
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
{{u}_{x}}=20m/s \\\ {{u}_{y}}=0 \\\ {{a}_{x}}=2m/{{s}^{2}},{{a}_{y}}=g=10m/{{s}^{2}} \\\ \Rightarrow 10=\dfrac{{{20}^{2}}{{\sin }^{2}}\theta }{20} \\\ \Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2} \\\ \Rightarrow \sin \theta =\dfrac{1}{\sqrt{2}} \\\ \therefore \theta ={{45}^{0}} \\\
Now let us find out the time taken to reach this height: Total time of flight is given by the formula $T=\dfrac{2u\sin \theta }{g}$
\Rightarrow T=\dfrac{2u\sin \theta }{g} \\\ \Rightarrow T=\dfrac{2\times 20\times \sin 45}{10} \\\ \Rightarrow T=\dfrac{4}{\sqrt{2}} \\\ \therefore T=2.82s \\\
Now we have the total time of flight and we now find out the horizontal distance covered,
$x={u_x}T+\dfrac{{a_x}{T^2}}{2} \\\ \Rightarrow x=20\times \cos 45\times 2.82+\dfrac{2\times {{2.82}^{2}}}{2} \\\ \Rightarrow x=39.88+7.95 \\\ \Rightarrow x=47.83m \\\ \therefore x\sim48m$
Thus, at the distance of 48 m from the front end of the train the ball lands.

Hence, option B is the correct answer.

Note: A projectile is any object thrown by the exertion of a force. Always in the formula the angle used is made with the vertical and if in the question it is given that angle is made with the vertical then we have to just subtract the given angle from ${{90}^{0}}$. The motion of the body has been split into its components and that is possible only because of the fact that acceleration was constant.