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NEET PHYSICSEasy

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is:

A

60°

B

15°

C

30°

D

45°

Step-by-Step Solution

  1. Velocity Components: A projectile launched with initial speed uu at an angle θ\theta has a horizontal velocity component ux=ucos⁡θu_x = u \cos \theta and a vertical component uy=usin⁡θu_y = u \sin \theta.
  2. Velocity at Maximum Height: At the maximum height, the vertical component of the velocity becomes zero (vy=0v_y = 0). Therefore, the speed of the projectile at maximum height is purely its horizontal component, which remains constant throughout the flight (neglecting air resistance) . vtop=ux=ucos⁡θv_{top} = u_x = u \cos \theta
  3. Given Condition: The speed at maximum height is half of the initial speed. vtop=u2v_{top} = \frac{u}{2}
  4. Calculation: Equating the expressions: ucos⁡θ=u2u \cos \theta = \frac{u}{2} cos⁡θ=12\cos \theta = \frac{1}{2} θ=cos⁡−1(0.5)=60∘\theta = \cos^{-1}(0.5) = 60^{\circ}
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