Two projectiles are thrown with the same rang | vedclass

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(D) For a given range $R$,the two angles of projection are $	heta$ and $(90^\circ - 	heta)$.The maximum height for angle $	heta$ is $h_1 = \frac{u^

Vedclass · Accepted Answer

$R = 4 \sqrt{h_1 h_2}$

Vedclass · Answer

(D) For a given range $R$,the two angles of projection are $	heta$ and $(90^\circ - 	heta)$.The maximum height for angle $	heta$ is $h_1 = \frac{u^2 \sin^2 	heta}{2g}$.The maximum height for angle $(90^\circ - 	heta)$ is $h_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Multiplying $h_1$ and $h_2$,we get:$h_1 h_2 = \frac{u^2 \sin^2 	heta}{2g} \cdot \frac{u^2 \cos^2 	heta}{2g} = \frac{u^4 \sin^2 	heta \cos^2 	heta}{4g^2}$.We know that the range $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.Squaring the range,$R^2 = \frac{4u^4 \sin^2 	heta \cos^2 	heta}{g^2}$.Comparing the two expressions,we see that $h_1 h_2 = \frac{R^2}{16}$.Therefore,$R^2 = 16 h_1 h_2$,which implies $R = 4 \sqrt{h_1 h_2}$.

Two projectiles are thrown with the same range $R$. If their maximum heights are $h_1$ and $h_2$ respectively,then what is the value of the range $R$?

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