For a projectile,the ratio of the maximum hei | vedclass

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(A) The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.We need to

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$5 : 4$

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(A) The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.We need to find the ratio $\frac{H}{T^2}$.$\frac{H}{T^2} = \frac{\frac{u^2 \sin^2 	heta}{2g}}{(\frac{2u \sin 	heta}{g})^2} = \frac{u^2 \sin^2 	heta}{2g} 	imes \frac{g^2}{4u^2 \sin^2 	heta}$.Simplifying the expression,we get $\frac{H}{T^2} = \frac{g}{8}$.Taking $g = 10 \ m/s^2$,the ratio is $\frac{10}{8} = \frac{5}{4}$.

For a projectile,the ratio of the maximum height reached to the square of the time of flight is:

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