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# 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
7 months ago

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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 $$h=\dfrac{2gu^2}{a^2}$$. · 7 months ago