# A Train is Moving along Straight Line with a Constant Acceleration a : Projectile Motion

**Problem:**
A train is moving along straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of 10 m/s at an angle 60° from horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train in m/s^{2} is, (IIT JEE 2011)

**Answer:** The answer is 5 m/s^{2}.

**Solution:**
Consider the motion in a frame attached to the ground. Let train's velocity at the time of projection of ball be $u_0$. The ball is projected with a velocity $u={10}\;\mathrm{m/s}$ (relative to the train) at an angle ${60}^\mathrm{o}$ from the horizontal (see figure).

Thus the horizontal and vertical component of the initial velocity of the ball are,
\begin{alignat}{2}
&u_{x}=u_0+u\cos 60, \nonumber\\
& u_{y}=u\sin 60. \nonumber
\end{alignat}
The time of flight is given by
\begin{align}
t&=\frac{2u_y}{g}
=\frac{2u\sin 60}{g}
=\sqrt{3}\;\mathrm{s}.
\end{align}
The distance travelled by the train and the ball in time $t$ are
\begin{align}
x_\text{train} &=u_0 t+\tfrac{1}{2}a t^2, \\
x_\text{ball} &=(u_0+u\cos 60) t.
\end{align}
Since the boy moves forward by 1.15 m, we have,
\begin{alignat}{2}
& x_\text{ball}=x_\text{train}+1.15, \nonumber\\
\label{gba:eqn:1}
& (u_0+u\cos 60) t=u_0 t+\tfrac{1}{2}a t^2+1.15.
\end{alignat}
Substitute $u={10}\;\mathrm{m/s}$ and $t=\sqrt{3}\;\mathrm{s}$ in first equation and then solve to get $a={5}\;\mathrm{m/s^2}$. I encourage you to solve this problem in a frame attached to the train.

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