The range of the projectile projected at an angle of 15° with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45° with horizontal, then its range will be

Solution

Physics sir · 3 steps

Same speed means the same u²/g, so the range follows sin 2θ alone.

1. Range formula. On flat ground the range is R = (u² sin 2θ)/g. The speed u is the same in both throws, so u²/g is a fixed number and only sin 2θ changes.
2. Use first throw. At 15°, sin 30° = 1/2, so 50 = u²/g × 1/2. That gives u²/g = 100 m.

   R₄₅/R₁₅ = (sin 90°)/(sin 30°) = 1/(1/2) = 2

3. So. R₄₅ = 2 × 50 = 100 m, which is option (b). This is also the maximum range, u²/g.

The idea to keep. When the launch speed stays the same, range is proportional to sin 2θ. Take the ratio, and remember that 45° gives the maximum range u²/g.

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