Question:
A boy is standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$ and the projection velocity in the vertical direction is $9.8$ $\mathrm{m} / \mathrm{s}$. How far behind the boy will the ball fall on the car?
Solution:
$=\frac{2 \mathrm{usin} \theta}{\mathrm{g}}$
$=\frac{2(9.8) \sin 90^{\prime}}{\mathrm{g}}$
$t=2 s e c$
For ball-car, along horizontal direction
$u_{\text {rel }}=0$
$a_{\text {rel }}=1$
$t=2$
Srel $=u_{r e \mid} t_{+} \frac{1}{2} a_{r e \mid} t^{2}$
$=0+^{\frac{1}{2}}(1)(2)^{2}$
$\mathrm{s}_{\mathrm{rel}}=2 \mathrm{~m}$