For a projectile the ratio of maximum height reached to the square of time of flight is $(g\ = \ 10\ ms^{- 1})$

Question

Dr. Abubacker · Accepted Answer

(c) Maximum height, $H = \frac{u^{2}\sin^{2}	heta}{2g}$
Time of flight, $T = \frac{2u\sin^{\ }	heta}{g}$
$	herefore\ \ \ \frac{H}{T^{2}} = \frac{u^{2}\sin^{2}{	heta/2g}}{4u^{2}\sin^{2}{	heta/g^{2}}} = \frac{g}{8} = \frac{10}{8} = \frac{5}{4}$

Q. For a projectile the ratio of maximum height reached to the square of time of flight is $(g\ = \ 10\ ms^{- 1})$

Explanation

Reviewed by Dr. Abubacker

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