A boy is riding on a flat car of a train moving with velocity $10 m{s}^{-1}$. The boy throws a ball which according to him makes angle $60°$ with the horizontal and in line with track. An observer on the ground observes the ball to rise vertically. He sees the ball to rise by __ m. (g=10ms^-2)

${\overrightarrow{v}}_{ball/train}={\overrightarrow{v}}_{ball}-{\overrightarrow{v}}_{train} \Rightarrow {\overrightarrow{v}}_{ball}={\overrightarrow{v}}_{ball/train}+{\overrightarrow{v}}_{train}$

Since ball moves in vertical direction w.r.t observer on ground, its velocity $x$-component direction will be zero.

Let the velocity of ball relative to train ${\overrightarrow{v}}_{BT}=\overrightarrow{v}$.

$$\begin{gathered} \therefore v_x=v \cos 60°-v_T=0 ......\left(1\right) \\ \text{and} v_y=v \sin 60° ......\left(2\right) \end{gathered}$$

From $\left(1\right)$, $v=\frac{v_T}{\cos 60°}=20 m{s}^{-1}$

and $v_y=20\times \sin 60°=10\sqrt{3}=10\sqrt{3} m{s}^{-1}$

$\therefore$ Maximum height reached by ball $=\frac{v_y^2}{2g}=\frac{{\left(10\sqrt{3}\right)}^2}{2\times 10}=15 m$.