A body is projected with speed $v\ ms^{- 1}$ at angle $ heta$. The kinetic energy at the highest point is half of the initial kinetic energy. The value of $ heta$ is

Question

A body is projected with speed $v\ ms^{- 1}$ at angle $	heta$. The kinetic energy at the highest point is half of the initial kinetic energy. The value of $	heta$ is

Dr. Abubacker · Accepted Answer

(b) Given $(KE)_{highest} = \frac{1}{2}(KE)$
$\frac{1}{2}mv^{2}\cos^{2}{	heta = \frac{1}{2}.\frac{1}{2}mv^{2}}$
$\cos^{2}{	heta = \frac{1}{2} \Rightarrow}\cos	heta = \sqrt{\frac{1}{2}}$
$\Rightarrow \ 	heta = 45{^\circ}$

Q. A body is projected with speed $v\ ms^{- 1}$ at angle $\theta$. The kinetic energy at the highest point is half of the initial kinetic energy. The value of $\theta$ is

Explanation

Reviewed by Dr. Abubacker

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