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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.

Respuesta :

Answer:

Explanation:

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

Time of fall  t = [tex]\sqrt{\frac{2h}{g} }[/tex]

Initial horizontal velocity = v ,

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

0 = ut - 1/2 a t²

0 = [tex]v\times\sqrt{\frac{2h}{g} } - \frac{1}{2}a \frac{2h}{g}[/tex]

v = 1/2 a [tex]\sqrt{\frac{2h}{g}[/tex]

h = [tex]\frac{2gv^2}{a^2}[/tex]

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