A projectile can have the same range for two | vedclass

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(A) The maximum height $h$ for a projectile is given by $h = \frac{u^2 \sin^2 	heta}{2g}$.For two angles of projection $	heta$ and $(90^\circ - 	he

Vedclass · Accepted Answer

$R = 4\sqrt{h_1 h_2}$

Vedclass · Answer

(A) The maximum height $h$ for a projectile is given by $h = \frac{u^2 \sin^2 	heta}{2g}$.For two angles of projection $	heta$ and $(90^\circ - 	heta)$,the range $R$ is the same,given by $R = \frac{u^2 \sin 2	heta}{g}$.Let $h_1 = \frac{u^2 \sin^2 	heta}{2g}$ and $h_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Multiplying $h_1$ and $h_2$:$h_1 h_2 = \left( \frac{u^2 \sin^2 	heta}{2g} ight) \left( \frac{u^2 \cos^2 	heta}{2g} ight) = \frac{u^4 \sin^2 	heta \cos^2 	heta}{4g^2}$.Since $R = \frac{u^2 (2 \sin 	heta \cos 	heta)}{g}$,we have $R^2 = \frac{u^4 (4 \sin^2 	heta \cos^2 	heta)}{g^2}$.Thus,$h_1 h_2 = \frac{u^4 \sin^2 	heta \cos^2 	heta}{4g^2} = \frac{R^2}{16}$.Therefore,$R^2 = 16 h_1 h_2$,which implies $R = 4 \sqrt{h_1 h_2}$.

$A$ projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are the maximum heights when the range in the two cases is $R$,then the relation between $R$,$h_1$,and $h_2$ is:

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