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A ball is projected with a velocity, $10 ms^{-1}$, at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be:
Explanation
$V_{Horizontal} = 10 \sin 60^{\circ} = 10 \times \sqrt{3}/2 = 5\sqrt{3}$ as there is zero acceleration along horizontal direction, at highest point we will have same horizontal velocity also there is no vertical velocity at highest point. so $|\vec{V}| = |\vec{V}_{Horizontal}| = 5\sqrt{3} m/s$