Problem
A student sits atop a platform a distance h above the ground. He throws a large object horizontally with a speed u. A wind blowing parallel to the ground gives the object a constant horizontal acceleration with magnitude a. This results in the object reaching the ground directly under the student. Determine the height h in terms of u, a and g. Ignore the effect of air resistance on the vertical motion.
Solution
For vertical motion,
$h = \frac{1}{2}g{t^2}$............*
For horizontal motion,
$0 = ut + \frac{1}{2}( - a){t^2}$
or $0 = u - \frac{1}{2}at$..............#
Putting $t = \sqrt {\frac{{2h}}{g}}$ from * in #,
$0 = u - \frac{1}{2}a\sqrt {\frac{{2h}}{g}}$
or $h = \frac{{2{u^2}g}}{{{a^2}}}$.