A projectile fired at 30° to the ground is observed to be at same height at time 3 s and 5 s after projection, during its flight. The speed of projection of the projectile is ___________ m s⁻¹.
(Given g=10 ms⁻² )

Solution

Physics sir · 3 steps

Equal heights at 3 s and 5 s means the top is at 4 s, so the whole flight lasts 8 s.

1. Use symmetry. The path is symmetric about the highest point. The projectile reaches the same height once going up and once coming down, at equal times before and after the top.
2. Time of flight. The top is midway between the two times, at (3+5)/2 = 4 s. So the total time of flight is T = 2 × 4 = 8 s. In general T = t₁ + t₂.

   T = (2u sin θ)/g ⇒ 8 = (2u × 1/2)/10 = u/10

3. So. u = 8 × 10 = 80 m s⁻¹, so the answer is 80.

The common slip: 40. Taking 4 s, the time to reach the top, as the full time of flight gives u = 40 m s⁻¹. The time of flight is twice that, 8 s.

The idea to keep. If a projectile is at the same height at times t₁ and t₂, then t₁ + t₂ is the full time of flight. You get it from symmetry alone, without writing the height equation.

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