NEET2026PhysicsChapterActual
A train of length L moving with a constant speed takes a time t₀ to completely cross a stationary pole. A projectile is fired with an initial speed equal to the speed of the train, at an angle to the horizontal. If the maximum height attained by the projectile is equal to the length of the train L , the angle of projection is given by
Options
- A= ⁻¹ ( 2g t₀^2 L )^ 1 2
- B= ⁻¹ ( 2g t₀^2 L )^ 1 2
- C= ⁻¹ ( g t₀^2 L )^ 1 2
- D= ⁻¹ ( L 2g t₀^2 )^ 1 2
Correct answer
A. = ⁻¹ ( 2g t₀^2 L )^ 1 2
Step-by-step solution
Let the speed of the train be v . Since the train of length L crosses a stationary pole in time t₀ , we have v = L t₀ . The projectile is fired with an initial speed u = v = L t₀ at an angle . The maximum height attained by the projectile is given by H = u^2 ^2 2g . Given that H = L , we can write: L = ( L t₀ )^2 ^2 2g L = L^2 ^2 2g t₀^2 ^2 = 2g t₀^2 L = ( 2g t₀^2 L )^ 1 2 = ⁻¹ ( 2g t₀^2 L )^ 1 2 Answer: = ⁻¹ ( 2g t₀^2 L )^ 1 2