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.

#Mechanics

Easy Math Editor

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Comments

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 $h=\dfrac{2gu^2}{a^2}$ .

A Former Brilliant Member - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$