JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 7)
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 -
1 : 4
1 : 2
4 : 3
4 : 1
Explanation
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$$ \mathrm{KE}_{\mathrm{in}}=\frac{1}{2} m v^{2} $$
$\mathrm{KE}_{\text {final }}=\frac{1}{2} m v^{2} \cos ^{2} 30^{\circ}=\frac{1}{2} m v^{2}\left(\frac{\sqrt{3}}{2}\right)^{2}$
$\frac{\mathrm{KE}_{\mathrm{in}}}{\mathrm{KE}_{\mathrm{f}}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{2}\left(\frac{3}{4}\right)}=\frac{4}{3}$
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