The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is √3/2 u. The time of flight of the projectile is :

Solution

Physics sir · 3 steps

At the top only the horizontal part u cos θ survives, so that speed tells you the angle.

1. Speed at the top. At the highest point the vertical velocity is zero. Gravity does not touch the horizontal velocity, so the speed there is u cos θ.
2. Find the angle. u cos θ = √3/2u gives cos θ = √3/2, so θ = 30° and sin θ = 1/2.

   T = (2u sin θ)/g = (2u × 1/2)/g = u/g

3. So. The time of flight is u/g, which is option (a).

The idea to keep. Speed at the highest point is always u cos θ, because the vertical part has gone to zero. Use it to get θ, then the vertical part u sin θ sets the time.

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