A stone projected with a velocity u at an ang | vedclass

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Question

(A) The maximum height reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first case,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.

Vedclass · Accepted Answer

$R = 4\sqrt{H_1 H_2}$

Vedclass · Answer

(A) The maximum height reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first case,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the second case,the angle is $(\frac{\pi}{2} - 	heta)$,so $H_2 = \frac{u^2 \sin^2(\frac{\pi}{2} - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Multiplying $H_1$ and $H_2$,we get $H_1 H_2 = \frac{u^4 \sin^2 	heta \cos^2 	heta}{4g^2} = \frac{u^4 (2 \sin 	heta \cos 	heta)^2}{16g^2} = \frac{(u^2 \sin 2	heta)^2}{16g^2}$.Since the horizontal range $R = \frac{u^2 \sin 2	heta}{g}$,we have $H_1 H_2 = \frac{R^2}{16}$.Therefore,$R^2 = 16 H_1 H_2$,which implies $R = 4\sqrt{H_1 H_2}$.

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