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Question

(A) For a given range $R$ and initial speed $u$,there are two possible angles of projection,$	heta$ and $(90^\circ - 	heta)$,that result in the same

Vedclass · Accepted Answer

$2R/g$

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(A) For a given range $R$ and initial speed $u$,there are two possible angles of projection,$	heta$ and $(90^\circ - 	heta)$,that result in the same horizontal range. The time of flight for these two angles are given by: $t_1 = \frac{2u \sin 	heta}{g}$ $t_2 = \frac{2u \sin(90^\circ - 	heta)}{g} = \frac{2u \cos 	heta}{g}$ The horizontal range $R$ is given by: $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$ Now,calculating the product $t_1t_2$: $t_1t_2 = \left( \frac{2u \sin 	heta}{g} ight) \left( \frac{2u \cos 	heta}{g} ight)$ $t_1t_2 = \frac{4u^2 \sin 	heta \cos 	heta}{g^2}$ Substituting the expression for $R$: $t_1t_2 = \frac{2}{g} \left( \frac{2u^2 \sin 	heta \cos 	heta}{g} ight) = \frac{2R}{g}$

$A$ shell is fired from a fixed artillery gun with an initial speed $u$ such that it hits the target on the ground at a distance $R$ from it. If $t_1$ and $t_2$ are the values of the time taken by it to hit the target in two possible ways,the product $t_1t_2$ is

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