Question

2. A projectile is fired with a velocity of 20ms⁻¹ at an angle of 40° to the horizontal. Determine the components of the velocity of the projectile at its maximum height.

Accepted Answer

At maximum height, the vertical component of velocity, vy = 0.
The horizontal component remains unchanged: vx = u cos40° = 20 × cos40° = 20 × 0.766 = 15.32 ms⁻¹

So at maximum height: horizontal component = 15.32 ms⁻¹, vertical component = 0 ms⁻¹.

WAEC Physics 2016 Theory — Question 2

Sign up free to unlock