2022 Motion in a PlaneProjectile Motion Medium Question 2 of 12 in Projectile Motion

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 :

A $10\ ms^{-1}$
B Zero
C $5\sqrt{3}\ ms^{-1}$
D $5\ ms^{-1}$

The answer is C.

Solution

At the highest point, only the horizontal component of velocity remains. Angle with the horizontal $\theta = 90^\circ - 60^\circ = 30^\circ$ $u_x = u\cos\theta = 10\cos 30^\circ = 5\sqrt{3}\ ms^{-1}$

Concept speed at highest point

Projectile Motion in past papers

12 questions from this chapter have appeared across 11 exam years.