Question 27

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 ?

A ball with kinetic energy $$KE$$ is projected at 60° from horizontal. At the highest point, only the horizontal component of velocity remains.

$$v_x = v\cos 60° = \frac{v}{2}$$

The kinetic energy at the highest point is

$$KE' = \frac{1}{2}mv_x^2 = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2}mv^2 = \frac{KE}{4}$$

Hence, the correct answer is Option 4: $$\frac{KE}{4}$$.

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