Question

A student sits atop a platform a distance h above the ground. He throws a pingpong ball horizontally with a speed v. However, a wind blowing parallel to the ground gives the pingpong ball a constant horizontal acceleration with magnitude a. This results in the pingpong ball reaching the ground directly under the student.
Determine the height h in terms of v, a, and g. This is a somewhat unphysical problem in that you can ignore the effect of air resistance on the vertical motion, but clearly the wind has a big affect on the horizontal motion.

Answer 1

Answer:

Explanation:

The pingpong ball reaches the ground directly under the student. That means net displacement is zero

Time of fall  t = $\sqrt{\frac{2h}{g} }$

Initial horizontal velocity = v ,

horizontal acceleration = - a ( deceleration because net displacement is zero)

0 = ut - 1/2 a t²

0 = $v\times\sqrt{\frac{2h}{g} } - \frac{1}{2}a \frac{2h}{g}$

v = 1/2 a $\sqrt{\frac{2h}{g}$

h = $\frac{2gv^2}{a^2}$