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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A train of length L moving with a constant speed takes a time t_0...

A train of length LL moving with a constant speed takes a time t0t_0 to completely cross a stationary pole. A projectile is fired with an initial speed equal to the speed of the train, at an angle θ\theta to the horizontal. If the maximum height attained by the projectile is equal to the length of the train LL, the angle of projection θ\theta is given by

Options

  1. A\theta = \sin^{-1}(2gt02L)12\left(\dfrac{2g t_0^2}{L}\right)^{\frac{1}{2}}
  2. B\theta = \cos^{-1}(2gt02L)12\left(\dfrac{2g t_0^2}{L}\right)^{\frac{1}{2}}
  3. C\theta = \sin^{-1}(gt02L)12\left(\dfrac{g t_0^2}{L}\right)^{\frac{1}{2}}
  4. D\theta = \sin^{-1}(L2gt02)12\left(\dfrac{L}{2g t_0^2}\right)^{\frac{1}{2}}

Correct answer

A. \theta = \sin^{-1}(2gt02L)12\left(\dfrac{2g t_0^2}{L}\right)^{\frac{1}{2}}

Step-by-step solution

Let the speed of the train be vv.

Since the train of length LL crosses a stationary pole in time t0t_0, we have v=Lt0v = \dfrac{L}{t_0}.

The projectile is fired with an initial speed u=v=Lt0u = v = \dfrac{L}{t_0} at an angle θ\theta.

The maximum height attained by the projectile is given by H=u2sin⁡2θ2gH = \dfrac{u^2 \sin^2 \theta}{2g}.

Given that H=LH = L, we can write:

L=(Lt0)2sin⁡2θ2gL = \dfrac{\left(\dfrac{L}{t_0}\right)^2 \sin^2 \theta}{2g}

L=L2sin⁡2θ2gt02L = \dfrac{L^2 \sin^2 \theta}{2g t_0^2}

sin⁡2θ=2gt02L\sin^2 \theta = \dfrac{2g t_0^2}{L}

sin⁡θ=(2gt02L)12\sin \theta = \left(\dfrac{2g t_0^2}{L}\right)^{\dfrac{1}{2}}

θ=sin⁡−1(2gt02L)12\theta = \sin^{-1}\left(\dfrac{2g t_0^2}{L}\right)^{\dfrac{1}{2}}

Answer: θ=sin⁡−1(2gt02L)12\theta = \sin^{-1}\left(\dfrac{2g t_0^2}{L}\right)^{\frac{1}{2}}
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