A ball is projected from the ground with a speed 15 ms⁻¹ at an angle θ with horizontal so that its range and maximum height are equal,
then 'tan θ' will be equal to :

Solution

Physics sir · 4 steps

Write R and H in terms of θ and set them equal. The speed cancels and only tan θ is left.

1. Write both formulas. For a projectile on flat ground, range is R = (u² sin 2θ)/g = (2u² sin θ cos θ)/g and maximum height is H = (u² sin² θ)/2g.
2. Set them equal. Put R = H. The common factor u²/g cancels, so the speed of 15 m s⁻¹ plays no part.

   2 sin θ cos θ = (sin² θ)/2

3. Solve for tan θ. Divide both sides by sin θ cos θ, which is not zero for a real projectile. This gives 2 = (tan θ)/2, so tan θ = 4.
4. So. tan θ = 4, which is option (d).

The idea to keep. In general H/R = (tan θ)/4. Remember this ratio and any question linking range and height becomes one line.

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