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(D) At point $A$, the kinetic energy is $(KE)_A = K = \frac{1}{2}mu^2$.At the highest point $B$, the vertical component of velocity is zero, so the ve

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$3$ : $4$

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(D) At point $A$, the kinetic energy is $(KE)_A = K = \frac{1}{2}mu^2$.At the highest point $B$, the vertical component of velocity is zero, so the velocity is $v_B = u \cos 60^{\circ} = \frac{u}{2}$.The kinetic energy at point $B$ is $(KE)_B = \frac{1}{2}m(\frac{u}{2})^2 = \frac{1}{4}(\frac{1}{2}mu^2) = \frac{K}{4}$.Since points $A$ and $C$ are at the same horizontal level, the speed at $C$ is equal to the speed at $A$. Thus, $(KE)_C = (KE)_A = K$.The difference in kinetic energies at points $B$ and $C$ is $|(KE)_C - (KE)_B| = |K - \frac{K}{4}| = \frac{3K}{4}$.The ratio of this difference to the kinetic energy at point $A$ is $\frac{3K/4}{K} = \frac{3}{4}$.

An object is projected with kinetic energy $K$ from a point $A$ at an angle $60^{\circ}$ with the horizontal. In the absence of air friction, what is the ratio of the difference in kinetic energies at points $B$ and $C$ to the kinetic energy at point $A$ (see figure)?

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