Two particles are projected from the same poi | vedclass

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(C) For the same range $R$,the angles of projection must be complementary,i.e.,$	heta$ and $(90^\circ - 	heta)$.The range $R$ is given by: $R = \fra

Vedclass · Accepted Answer

$R^2 = 16 h_1h_2$

Vedclass · Answer

(C) For the same range $R$,the angles of projection must be complementary,i.e.,$	heta$ and $(90^\circ - 	heta)$.The range $R$ is given by: $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin	heta \cos	heta}{g}$.The maximum heights for the two particles are:$h_1 = \frac{u^2 \sin^2	heta}{2g}$$h_2 = \frac{u^2 \sin^2(90^\circ - 	heta)}{2g} = \frac{u^2 \cos^2	heta}{2g}$Multiplying $h_1$ and $h_2$:$h_1 h_2 = \left( \frac{u^2 \sin^2	heta}{2g} ight) \left( \frac{u^2 \cos^2	heta}{2g} ight) = \frac{u^4 \sin^2	heta \cos^2	heta}{4g^2}$$h_1 h_2 = \frac{1}{16} \left( \frac{4u^4 \sin^2	heta \cos^2	heta}{g^2} ight) = \frac{1}{16} R^2$Therefore,$R^2 = 16 h_1 h_2$.

Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$,but different maximum heights,$h_1$ and $h_2$. Which of the following is correct?

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