Question 92: Work, Energy and Power & Circular Motion

Question

A stone is projected at angle $30^{\circ}$ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be -

Accepted Answer

$
\mathrm{KE}_{\mathrm{in}}=\frac{1}{2} m v^{2}
$
$\mathrm{KE}_{	ext {final }}=\frac{1}{2} m v^{2} \cos ^{2} 30^{\circ}=\frac{1}{2} m v^{2}\left(\frac{\sqrt{3}}{2}ight)^{2}$
$\frac{\mathrm{KE}_{\mathrm{in}}}{\mathrm{KE}_{\mathrm{f}}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{2}\left(\frac{3}{4}ight)}=\frac{4}{3}$

Answer

1 : 4

Answer

1 : 2

Answer

4 : 3

Answer

4 : 1

Work, Energy and Power & Circular Motion

A stone is projected at angle 30^{\circ} to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be -

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