Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T₁ and T₂ are the total flying times of first and second ball, respectively, then the ratio of T₁ and T₂ is

Solution

Physics sir · 3 steps

Height and flight time both depend only on the vertical velocity, so link them and take a square root.

1. Vertical part only. Mass plays no role in projectile motion. Only the vertical velocity u_y = u sin θ fixes how high the ball goes and how long it stays up.
2. Two formulas. Maximum height is H = u_y²/2g and total time of flight is T = 2u_y/g. So H ∝ u_y² while T ∝ u_y, which means T ∝ √H.

   T₁/T₂ = √(H₁/H₂) = √8 = 2√2

3. So. T₁ : T₂ = 2√2 : 1, which is option (c).

The idea to keep. Height goes as the square of the vertical velocity, but flight time goes as the vertical velocity itself. So the time ratio is always the square root of the height ratio.

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