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A boy is standing on a long railroad car throws a ball straight upwards.

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 $=u_{r e \mid} t_{+} \frac{1}{2} a_{r e \mid} t^{2}$

$=0+^{\frac{1}{2}}(1)(2)^{2}$

$\mathrm{s}_{\mathrm{rel}}=2 \mathrm{~m}$

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