Question

A train is moving along a straight line with a constant acceleration $a$. A boy standing in the train throws a ball forward with a speed a of $10m{s}^{-1}$, at an angle of ${60}^o$ to the horizontal. The body has to move forward by $1.15m$ inside the train to catch the ball back to the initial height. The acceleration of the train, in $m{s}^{-2}$, is

Answer 1

Initial velocity of ball= $u_o$

Time of flight= $\frac{2u_o\sin \theta }{g}=\frac{2\times 10\sin {60}^o}{10}=\frac{20\times \frac{\sqrt{3}}{2}}{10}=\sqrt{3}sec$

Now, $x=A{u}_o\cos \theta t+\frac{1}{2}a{t}^{2}A$

$\Rightarrow x=10\times cos{60}^o\times \sqrt{3}+\frac{1}{2}a(\sqrt{3}{)}^2$

$\Rightarrow x=A5\sqrt{3}+\frac{3a}{2}A$

Since boy has to move $1.5m$ forward to catch the ball;

So, x=1.15

$\Rightarrow 5\sqrt{3}+\frac{3a}{2}=1.15$

$\Rightarrow 3a=-15.02$ or $a=-5m/se{c}^2$