A ball is projected with a velocity, $10 ext{ ms}^{-1}$, at an angle of $60^\circ$ with the vertical directi

Question

A ball is projected with a velocity, $10 	ext{ ms}^{-1}$, at an angle of $60^\circ$ with the vertical direction. Its speed at the highest point of its trajectory will be:

Accepted Answer

The angle with the vertical is $60^\circ$, so the angle with the horizontal is $	heta = 90^\circ - 60^\circ = 30^\circ$. At the highest point, the vertical component of velocity is zero, and the speed is equal to the horizontal component: $v = u \cos(	heta) = 10 \cos(30^\circ) = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3} 	ext{ ms}^{-1}$.

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