JEE Main 2025 · 7 April · morning · Single correct · +4 / −1
Two projectiles are fired from ground with same initial speeds from same point at angles (45°+ α) and (45°-α) with horizontal direction. The ratio of their times of flights is
- (a) (1+tan α)/(1-tan α)
- (b) (1+sin 2 α)/(1-sin 2 α)
- (c) (1-tan α)/(1+tan α)
- (d) 1
Answer and solution
Answer (a) (1+tan α)/(1-tan α)
Solution
Physics sir · 4 steps
Time of flight goes as sin θ, so just divide the two sines and simplify.
- Time of flight. For a projectile fired from the ground, T = (2u sin θ)/g. Both have the same u and g, so only sin θ changes.
T₁/T₂ = sin(45°+α)/sin(45°-α)
- Expand the sines. sin(45°±α) = 1/√2(cos α ± sin α). The 1/√2 cancels, leaving (cos α+sin α)/(cos α-sin α).
- Divide by cos. Divide the top and bottom by cos α to get (1+tan α)/(1-tan α).
- So. The ratio of times of flight is (1+tan α)/(1-tan α), which is option (a).
The idea to keep. Complementary angles give equal ranges, but not equal times. Time goes as sin θ and maximum height goes as sin² θ, so the steeper shot stays up longer.
Answer from the NTA final answer key. Solution by PrepWiser.