- 2018: A particle is projected with velocity u at an angle $\theta$ with the horizontal. The average velocity of the particle between the time it reaches the maximum height and the time it reaches the ground is:
(A) $\frac{u}{2}$ (B) $\frac{u}{\sqrt{2}}$ (C) $\frac{u}{\sqrt{3}}$ (D) $\frac{u}{3}$
The answer is (D)
The average velocity is the total displacement divided by the total time. The total displacement is the distance from the point of projection to the ground, which is $2u\sin\theta$. The total time is the time it takes to reach the maximum height, which is $\frac{u\sin\theta}{g}$, plus the time it takes to fall from the maximum height to the ground, which is $\frac{u\sin\theta}{g}$. So the total time is $\frac{2u\sin\theta}{g}$. The average velocity is therefore $\frac{2u\sin\theta}{2u\sin\theta/g} = \frac{u}{g}$
