2016: Problem 16

Topic: Kinematics
Concepts: Projectile motion, Relative velocity

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Solution:

We go to the truck frame where it is at rest and the ball is moving at $v=8\,\mathrm{m/s}$ to the left. To drop by distance $h$, it takes time

$$h=\frac{1}{2}gt^2$$

$$t=\sqrt{\frac{2h}{g}}$$

In this time, the ball travels a distance

$$d=vt=v\sqrt{\frac{2h}{g}}=(8\,\mathrm{m/s})\sqrt{\frac{2(1.25\,\mathrm{m})}{(10\,\mathrm{m/s^2})}}=4\,\mathrm{m}$$

to the left so the answer is C.

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