For a projectile, the ratio of maximum height reached to the square of flight time is $(g = 10\ ms^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(a) $H = \frac{u^{2}\sin^{2}	heta}{2g}\ $and $T = \frac{2u\sin	heta}{g}$
So $\frac{H}{T^{2}} = \frac{u^{2}\sin^{2}	heta/2g}{4u^{2}\sin^{2}	heta/g^{2}} = \frac{g}{8} = \frac{5}{4}$

Q. For a projectile, the ratio of maximum height reached to the square of flight time is $(g = 10\ ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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