A particle is projected with velocity u so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as (n u²)/(25 g), where value of n is: (Given, ' g ' is the acceleration due to gravity.)

Solution

Physics sir · 4 steps

Set R equal to 3H, and the condition gives you tan θ. A 3-4-5 triangle then does the rest.

1. Write both formulas. For launch angle θ, the range is R = (u² sin 2θ)/g = (2u² sin θ cos θ)/g. The maximum height is H = (u² sin² θ)/2g.

   (2u² sin θ cos θ)/g = 3 × (u² sin² θ)/2g ⇒ tan θ = 4/3

2. Find the angle. Cancel (u² sin θ)/g from both sides to get 2 cos θ = 3/2 sin θ, so tan θ = 4/3. This is a 3-4-5 triangle, so sin θ = 4/5 and cos θ = 3/5.
3. Put into range. sin 2θ = 2 × 4/5 × 3/5 = 24/25. So R = 24u²/25g.
4. So. Comparing with nu²/25g gives n = 24, which is option (d).

The idea to keep. Any condition linking R and H fixes the angle. Use R/H = 4/(tan θ) to find tan θ, then read off sin 2θ from the right triangle.

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