A ball is projected with speed 20 m/s at an angle of 30° with the horizontal. What is the maximum height reached? (Take g = 10 m/s²)

Solution

Physics sir · 3 steps

Only the vertical part of the velocity decides how high the ball goes, so square u sin θ.

1. Vertical component. The vertical part of the velocity is u sin 30° = 20 × 1/2 = 10 m s⁻¹. Gravity slows only this part, and at the top it becomes zero.

   H = (u² sin² θ)/2g

2. Put in numbers. H = (10)²/(2 × 10) = 100/20 = 5 m.
3. So. The maximum height is 5 m, which is option (a).

The idea to keep. For maximum height, think of the projectile as a ball thrown straight up with speed u sin θ. The horizontal motion plays no part.

Answer from the NTA final answer key. Solution by PrepWiser.