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Question

(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that $(H_{\max})_1 = 8 	imes (H_{\max})_2$,we have:$\frac

Vedclass · Accepted Answer

$2 \sqrt{2} : 1$

Vedclass · Answer

(A) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that $(H_{\max})_1 = 8 	imes (H_{\max})_2$,we have:$\frac{u^2 \sin^2 	heta_1}{2g} = 8 	imes \frac{u^2 \sin^2 	heta_2}{2g}$.Simplifying this,we get $\sin^2 	heta_1 = 8 \sin^2 	heta_2$,which implies $\sin 	heta_1 = \sqrt{8} \sin 	heta_2 = 2\sqrt{2} \sin 	heta_2$.The time of flight is given by $T = \frac{2u \sin 	heta}{g}$.Therefore,the ratio of the times of flight is $\frac{T_1}{T_2} = \frac{2u \sin 	heta_1 / g}{2u \sin 	heta_2 / g} = \frac{\sin 	heta_1}{\sin 	heta_2}$.Substituting the value of $\sin 	heta_1$,we get $\frac{T_1}{T_2} = \frac{2\sqrt{2} \sin 	heta_2}{\sin 	heta_2} = 2\sqrt{2}$.Thus,the ratio is $2\sqrt{2} : 1$.

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