The velocity at the maximum height of a proje | vedclass

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Question

(A) At the maximum height,the vertical component of velocity is zero,so the velocity of the projectile is equal to its horizontal component,$v_x = u \

Vedclass · Accepted Answer

$\frac{\sqrt{3} u^{2}}{2 g}$

Vedclass · Answer

(A) At the maximum height,the vertical component of velocity is zero,so the velocity of the projectile is equal to its horizontal component,$v_x = u \cos 	heta$.Given that this velocity is half of the initial velocity $u$,we have $u \cos 	heta = \frac{u}{2}$.This implies $\cos 	heta = \frac{1}{2}$,so the angle of projection is $	heta = 60^\circ$.The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$.Substituting $	heta = 60^\circ$,we get $R = \frac{u^2 \sin(2 	imes 60^\circ)}{g} = \frac{u^2 \sin(120^\circ)}{g}$.Since $\sin(120^\circ) = \frac{\sqrt{3}}{2}$,the range is $R = \frac{\sqrt{3} u^2}{2g}$.

The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is

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