A body of mass $m$ is projected with initial velocity $v_0$ at an angle $\alpha$ above the horizonta

Question

A body of mass $m$ is projected with initial velocity $v_0$ at an angle $\alpha$ above the horizontal. Neglecting air resistance and taking the gravitational acceleration to be $g$ downwards, which of the following statements regarding the power developed by the gravitational force are correct?

Professor PC Thomas Classes · Answer

The gravitational force is $\vec{F} = -mg \hat{j}$. The velocity vector at time $t$ is $\vec{v}(t) = (v_0 \cos\alpha)\hat{i} + (v_0 \sin\alpha - gt)\hat{j}$. The instantaneous power is $P(t) = \vec{F} \cdot \vec{v} = -mg(v_0 \sin\alpha - gt) = -mgv_0 \sin\alpha + mg^2 t$, confirming A. At the top of the trajectory ($t = v_0 \sin\alpha / g$), $\vec{v}$ is horizontal while $\vec{F}$ is vertical, so $P = 0$, which is the minimum magnitude, confirming D is incorrect. The ascent time is $T_1 = v_0 \sin\alpha / g$. The mean power during ascent is $\langle P \rangle_{ascent} = \frac{1}{T_1} \int_0^{T_1} P(t) dt = \frac{\Delta K_{ascent}}{T_1} = \frac{- \frac{1}{2} m (v_0 \sin\alpha)^2}{v_0 \sin\alpha / g} = -\frac{1}{2} mgv_0 \sin\alpha$, confirming B. Over the full symmetric flight, net change in kinetic energy is zero, so total mean power is zero, confirming C.