An object is projected with kinetic energy $K$ from a point $A$ at an angle $60^{\circ}$ with the ho

Question

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

Professor PC Thomas Classes · Answer

Let the initial velocity be $v_0$. The initial kinetic energy at the launch point A is $K_A = K = \frac{1}{2}mv_0^2$. At the highest point B, the vertical velocity is zero, and only the horizontal component $v_x = v_0\cos(60^\circ) = \frac{v_0}{2}$ remains. Thus, $K_B = \frac{1}{2}m(\frac{v_0}{2})^2 = \frac{K}{4}$. Assuming C is the landing point on the same horizontal level, by energy conservation, $K_C = K_A = K$. The difference in kinetic energies at points C and B is $K_C - K_B = K - \frac{K}{4} = \frac{3K}{4}$. The ratio to the kinetic energy at A is $\frac{3K/4}{K} = 3:4$.