Question 232 of 308

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 :

A$1:2$B$1:4$C$4:1$D$4:3$

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Answer: (D) $4:3$

Let stone is projected with initial velocity $v$.

Kinetic energy of the stone at point of projection is $K{E}_{1}=\dfrac{1}{2}m{v}^{2}$.
Now, velocity of stone at maximum height is $=v\cos 30^{\circ}$, so kinetic energy at that point is $K{E}_{2}=\dfrac{1}{2}m{\left(v\cos 30^{\circ}\right)}^{2}$
Hence, the ratio $\dfrac{K{E}_{1}}{K{E}_{2}}=\dfrac{\dfrac{1}{2}m{v}^{2}}{\dfrac{1}{2}m{\left(v\cos 30^{\circ}\right)}^{2}}=\dfrac{4}{3}$