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

<p>Let's break down the problem step by step:</p> <p><p>The ball is projected with kinetic energy </p> <p>$\mathrm{KE} = \frac{1}{2} m v^2,$</p> <p>where $v$ is the initial speed.</p></p> <p><p>The ball is launched at an angle of $60^\circ$. So, its initial velocity components are:</p></p> <p><p>Horizontal: $v_x = v \cos(60^\circ) = \frac{v}{2}$</p></p> <p><p>Vertical: $v_y = v \sin(60^\circ)$</p></p> <p><p>At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., $v_y = 0$), but the horizontal component remains unchanged.</p></p> <p><p>Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:</p> <p>$\mathrm{KE}_{\text{top}} = \frac{1}{2} m v_x^2$</p></p> <p><p>Substitute $v_x = \frac{v}{2}$:</p> <p>$$\mathrm{KE}_{\text{top}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2$$</p></p> <p><p>Notice that the initial kinetic energy is </p> <p>$\mathrm{KE} = \frac{1}{2} m v^2.$ </p> <p>So we can write:</p> <p>$$\mathrm{KE}_{\text{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} \mathrm{KE}$$</p></p> <p>Thus, the kinetic energy at the highest point is $\frac{\mathrm{KE}}{4}$.</p> <p>Answer: Option C.</p>