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

1. To determine the kinetic energy of a ball at the highest point of its flight, we begin by analyzing the components of motion:
   • A ball is projected at an angle of \(60^\circ\) from the horizontal.
   • The kinetic energy \(KE\) of the ball at the point of projection can be associated with its initial velocity \(v_0\).
2. The kinetic energy at any point is given by the equation: \(KE = \frac{1}{2} m v^2\).
3.  At the highest point of the flight:
   • The ball's velocity is entirely horizontal, as the vertical component of velocity becomes zero.
   • The initial velocity can be decomposed into horizontal and vertical components:
     • Horizontal component: \(v_0 \cos(60^\circ)\), which simplifies to \(\frac{v_0}{2}\).
     • Vertical component: \(v_0 \sin(60^\circ)\).
4. At the highest point, the kinetic energy is due to the horizontal velocity only:
   • The horizontal velocity remains constant as \(\frac{v_0}{2}\).
5. The initial kinetic energy \(KE\) is:
   • \(\frac{1}{2} m v_0^2\)
6. Therefore, the kinetic energy of the ball at the highest point of its flight is \(\frac{KE}{4}\).

The correct option is, therefore, \(\frac{KE}{4}\).

Answer 2

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: \[ KE_{\text{highest}} = \frac{1}{2} m (v \cos(60^\circ))^2 = \frac{1}{2} m v^2 \cdot \frac{1}{4} = \frac{KE}{4}. \]