For a projectile, the ratio of maximum height | vedclass

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Question

(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}}{{

Vedclass · Accepted Answer

$5:4$

Vedclass · Answer

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

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

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