Question 32: Work, Energy and Power & Circular Motion

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 ball at the highest point of its flight?

Accepted Answer

Let's break down the problem step by step:
The ball is projected with kinetic energy
$$\mathrm{KE} = \frac{1}{2} m v^2,$$
where $$v$$ is the initial speed.
The ball is launched at an angle of $$60^\circ$$. So, its initial velocity components are:
Horizontal: $$v_x = v \cos(60^\circ) = \frac{v}{2}$$
Vertical: $$v_y = v \sin(60^\circ)$$
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.
Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:
$$\mathrm{KE}_{	ext{top}} = \frac{1}{2} m v_x^2$$
Substitute $$v_x = \frac{v}{2}$$:
$$\mathrm{KE}_{	ext{top}} = \frac{1}{2} m \left(\frac{v}{2}ight)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2$$
Notice that the initial kinetic energy is
$$\mathrm{KE} = \frac{1}{2} m v^2.$$
So we can write:
$$\mathrm{KE}_{	ext{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2ight) = \frac{1}{4} \mathrm{KE}$$
Thus, the kinetic energy at the highest point is $$\frac{\mathrm{KE}}{4}$$.
Answer: Option C.

Answer

$\frac{(\mathrm{KE})}{2}$

Answer

$\frac{(\mathrm{KE})}{8}$

Answer

$\frac{(\mathrm{KE})}{4}$

Answer

$\frac{(\mathrm{KE})}{16}$

Work, Energy and Power & Circular Motion

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?

Choose the correct answer: