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Question

(A) Let the initial velocity be $u$ and the angle of projection be $	heta$. At maximum height,the vertical component of velocity is $0$,so the veloci

Vedclass · Accepted Answer

$\frac{u^2}{g}$

Vedclass · Answer

(A) Let the initial velocity be $u$ and the angle of projection be $	heta$. At maximum height,the vertical component of velocity is $0$,so the velocity is only the horizontal component,$v_x = u \cos 	heta$. Given that $v_x = \frac{u}{\sqrt{2}}$,we have $u \cos 	heta = \frac{u}{\sqrt{2}}$. This implies $\cos 	heta = \frac{1}{\sqrt{2}}$,so $	heta = 45^o$. The formula for the range $R$ of a projectile is $R = \frac{u^2 \sin(2	heta)}{g}$. Substituting $	heta = 45^o$,we get $R = \frac{u^2 \sin(2 	imes 45^o)}{g} = \frac{u^2 \sin(90^o)}{g}$. Since $\sin(90^o) = 1$,the range is $R = \frac{u^2}{g}$.

If the velocity of a projectile at its maximum height is $\frac{1}{\sqrt{2}}$ times its initial velocity,what is its range?

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