Two projectiles P₁ and P₂ thrown with speed in the ratio √3 : √2, attain the same height during their motion. If P₂ is thrown at an angle of 60° with the horizontal, the angle of projection of P₁ with horizontal will be :

Solution

Physics sir · 3 steps

Same maximum height means the vertical parts of the two launch velocities are equal. Match those.

1. Height needs vertical speed. The maximum height is H = (u² sin² θ)/2g. It depends only on the vertical component u sin θ, so equal heights mean equal u sin θ.

   u₁ sin θ₁ = u₂ sin θ₂

2. Put in numbers. Take u₁ = √3 and u₂ = √2, with θ₂ = 60°. Then √3 sin θ₁ = √2 × √3/2, so sin θ₁ = √2/2 = 1/√2.
3. So. sin θ₁ = 1/√2 gives θ₁ = 45°, which is option (c).

The idea to keep. Maximum height and time of flight depend only on u sin θ. Range depends on both components. When two heights are equal, set the vertical components equal and solve.

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