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:
$\underline{2 u \sin \theta}$
Time of flight for ball $=$ g
$=\frac{2(9.8) \sin 9 \overline{0}^{-}}{\text {E }}$
$t=2 \sec$
For ball-car, along horizontal direction
$u_{r e l}=0$
$a_{r e l}=1$
$t=2$
$\mathrm{S}_{\mathrm{rel}}=\mathrm{U}_{\mathrm{rel}} t_{+} \frac{1}{2} \mathrm{a}_{\mathrm{rel}} t^{2}$
$=0+^{\frac{1}{2}}(1)(2)^{2}$
$\mathrm{s}_{\mathrm{rel}}=2 \mathrm{~m}$