A projectile can have the same range R for tw | vedclass

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(B) For a projectile with initial velocity $u$ and projection angle $	heta$,the range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$.The time of f

Vedclass · Accepted Answer

$t_1 t_2 \propto R$

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(B) For a projectile with initial velocity $u$ and projection angle $	heta$,the range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$.The time of flight $t$ is given by $t = \frac{2u \sin	heta}{g}$.For the same range $R$,the two angles of projection are $	heta$ and $(90^\circ - 	heta)$.Let $t_1 = \frac{2u \sin	heta}{g}$ and $t_2 = \frac{2u \sin(90^\circ - 	heta)}{g} = \frac{2u \cos	heta}{g}$.Multiplying $t_1$ and $t_2$ gives:$t_1 t_2 = \left( \frac{2u \sin	heta}{g} ight) \left( \frac{2u \cos	heta}{g} ight) = \frac{4u^2 \sin	heta \cos	heta}{g^2}$.Using the identity $\sin(2	heta) = 2 \sin	heta \cos	heta$,we get:$t_1 t_2 = \frac{2u^2 \sin(2	heta)}{g^2}$.Since $R = \frac{u^2 \sin(2	heta)}{g}$,we can substitute $u^2 \sin(2	heta) = Rg$:$t_1 t_2 = \frac{2(Rg)}{g^2} = \frac{2R}{g}$.Since $2$ and $g$ are constants,$t_1 t_2 \propto R$.

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

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