The initial speed of a projectile fired from | vedclass

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(B) At the highest point of a projectile's trajectory,the vertical component of velocity is zero,and the speed is equal to the horizontal component of

Vedclass · Accepted Answer

$\frac{u}{g}$

Vedclass · Answer

(B) At the highest point of a projectile's trajectory,the vertical component of velocity is zero,and the speed is equal to the horizontal component of velocity,which is $v_x = u \cos 	heta$.Given that the speed at the highest point is $\frac{\sqrt{3}}{2} u$,we have $u \cos 	heta = \frac{\sqrt{3}}{2} u$.This implies $\cos 	heta = \frac{\sqrt{3}}{2}$,so $	heta = 30^{\circ}$.The time of flight $T$ is given by the formula $T = \frac{2u \sin 	heta}{g}$.Substituting $	heta = 30^{\circ}$,we get $T = \frac{2u \sin 30^{\circ}}{g} = \frac{2u (1/2)}{g} = \frac{u}{g}$.

The initial speed of a projectile fired from the ground is $u$. At the highest point during its motion,the speed of the projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:

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