A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight? [2025]

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Solution

Q.
A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight [2025]

1 $\frac{KE}{8}$

2 $\frac{KE}{4}$

3 $\frac{KE}{16}$

4 $\frac{KE}{2}$

Ans.
(2)

Initial kinetic energy, $K . E = \frac{1}{2} m u^{2}$

Speed at highest point, $v = u \cos 60 ° = \frac{u}{2}$

$∴ {KE}_{2} = \frac{1}{2} m {(\frac{u}{2})}^{2} = \frac{1}{4} \times \frac{1}{2} m u^{2} = \frac{KE}{4}$

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