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?
Select the correct option:
A
$\frac{(\mathrm{KE})}{8}$
B
$\frac{(\mathrm{KE})}{2}$
C
$\frac{(\mathrm{KE})}{4}$
D
$\frac{(\mathrm{KE})}{16}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Initial K.E,
K.E. $=\frac{1}{2} \mathrm{mu}^{2}$
Speed at highest point
$\mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2}$
$\therefore \mathrm{KE}_{2}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^{2}$
$=\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^{2}=\frac{\mathrm{KE}}{4}$
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