Two bodies are projected from ground with same speeds 40 ms⁻¹ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of 60°, with horizontal then sum of the maximum heights, attained by the two projectiles, is m. (Given g=10 ms⁻² )

Solution

Physics sir · 3 steps

Same speed, same range means the angles are complementary, so the other body went at 30°.

1. Find the other angle. Range is R = (u² sin 2θ)/g. Since sin 2θ = sin(180° - 2θ), angles θ and 90° - θ give the same range. So the second body was projected at 90° - 60° = 30°.

   H₁ + H₂ = (u² sin² 60°)/2g + (u² sin² 30°)/2g = u²/2g(sin² 60° + cos² 60°) = u²/2g

2. Put in numbers. u²/2g = 40²/(2 × 10) = 1600/20 = 80 m. As a check, the individual heights are 60 m at 60° and 20 m at 30°.
3. So. The sum of the maximum heights is 80 m.

The common slip: 60. Finding the maximum height for the 60° body alone gives 60, not 80. Add the 20 m from the 30° body as well.

The idea to keep. Equal range at equal speed means complementary angles. Then sin² θ + cos² θ = 1 gives H₁ + H₂ = u²/2g, whatever the angle.

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