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.

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TopNewestSince there is no initial velocity, \(h=\dfrac{1}{2}gt^2\)(1).

Since the object reaches directly under the student, \(s=0=ut+\dfrac{1}{2}(-a)t^2\)(acceleration is negative as it is in the opposite direction) or \(0=u-\dfrac{1}{2}at\) or \(t=\dfrac{2u}{a}\)(2).

Substituting the value of (2) into (1) we get \(h=\dfrac{2gu^2}{a^2}\). – Brilliant Member · 1 year, 6 months ago

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