For a projectile,the ratio of maximum height | vedclass

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Question

(A) The maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.The time of flight $T$ is given by $T = \frac{2u \si

Vedclass · Accepted Answer

$5:4$

Vedclass · Answer

(A) The maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.Squaring the time of flight,we get $T^2 = \frac{4u^2 \sin^2 	heta}{g^2}$.Now,the ratio of maximum height to the square of flight time is $\frac{H}{T^2} = \frac{u^2 \sin^2 	heta / 2g}{4u^2 \sin^2 	heta / g^2}$.Simplifying this expression: $\frac{H}{T^2} = \frac{u^2 \sin^2 	heta}{2g} 	imes \frac{g^2}{4u^2 \sin^2 	heta} = \frac{g}{8}$.Given $g = 10 \ m/s^2$,we have $\frac{H}{T^2} = \frac{10}{8} = \frac{5}{4}$.

For a projectile,the ratio of maximum height reached to the square of flight time is $(g = 10 \ m/s^2)$.

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