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 apex of its trajectory, we first analyze the motion's components:
   • A ball is launched at an angle of \(60^\circ\) with respect to the horizontal.
   • The kinetic energy \(KE\) at the launch point is derived from its initial velocity \(v_0\).
2. The kinetic energy at any point is calculated using the formula: \(KE = \frac{1}{2} m v^2\).
3. At the highest point of the flight:
   • The ball's velocity is exclusively horizontal, as its vertical velocity component is zero.
   • The initial velocity can be resolved 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 apex, kinetic energy is solely attributable to the horizontal velocity:
   • The horizontal velocity remains constant at \(\frac{v_0}{2}\).
5. The initial kinetic energy \(KE\) is given by:
   • \(\frac{1}{2} m v_0^2\)
6. Consequently, the kinetic energy of the ball at the highest point of its trajectory is \(\frac{KE}{4}\).

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