Waste less time on Facebook — follow Brilliant.

My first classical note

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.

Note by Lakshya Sinha
1 year, 6 months ago

No vote yet
1 vote


Sort by:

Top Newest

Since 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

Log in to reply


Problem Loading...

Note Loading...

Set Loading...