The maximum height attained by a projectile when thrown at an angle $ heta$ with the horizontal is found to be half the horizontal range. Then $ heta$ is equal to

Question

The maximum height attained by a projectile when thrown at an angle $	heta$ with the horizontal is found to be half the horizontal range. Then $	heta$ is equal to

Dr. Abubacker · Accepted Answer

(a) $Maximum\ height,\ H = \frac{u^{2}\sin^{2}	heta}{2g}$
$Range,\ \ R = \frac{u^{2}sin2	heta}{g}$
$Given,\ \ H = \frac{R}{2}$
$	herefore\ \frac{u^{2}\sin^{2}	heta}{2g} = \frac{u^{2}2sin	heta cos	heta}{2g}$
$or\ \ sin	heta = 2cos	heta$
$or\ \ tan	heta = 2$
$or\ \ 	heta = 	an^{- 1}(2)$

Q. The maximum height attained by a projectile when thrown at an angle $\theta$ with the horizontal is found to be half the horizontal range. Then $\theta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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