Question

A ball having kinetic energy \( KE \), is projected at an angle of \( 60^\circ \) from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

• A. \( \frac{KE}{8} \)
• B. \( \frac{KE}{4} \)
• C. \( \frac{KE}{16} \)
• D. \( \frac{KE}{2} \)

Hint:

At the highest point of projectile motion, the kinetic energy is reduced because only the horizontal velocity contributes to the energy.

Answer 1

The Correct Option is B

At the highest point of the flight, the vertical component of the velocity becomes zero. Therefore, the kinetic energy at the highest point is only due to the horizontal component of the velocity. The initial kinetic energy \( KE \) is the sum of the horizontal and vertical components: \[ KE = \frac{1}{2} m v^2, \] where \( v \) is the initial velocity. At the highest point, only the horizontal component of the velocity remains, which is \( v \cos(60^\circ) \). Thus, the kinetic energy at the highest point is: \[ KE_{\text{highest}} = \frac{1}{2} m (v \cos(60^\circ))^2 = \frac{1}{2} m v^2 \cdot \frac{1}{4} = \frac{KE}{4}. \]