Question

A ball is thrown such that it strikes the ground with velocity $v_0$ making an angle $\theta$ with the vertical. The height to which it rises after the collision (if the coefficient of restitution is $e$) is

• A. $\frac{e^2{v}_0^2{\text{cos}}^2\theta }{2g}$
• B. $\frac{e^2{v}_0^2{\text{sin}}^2\theta }{2g}$
• C. $\frac{e^2{v}_0^2\text{tan}\theta }{2g}$
• D. $\frac{e^2{v}_0^2{\text{cos}}^2\theta }{g}$

Answer 1

The correct option is A

$\frac{e^2{v}_0^2{\text{cos}}^2\theta }{2g}$

$e=-\left[\frac{v_y}{v_{0y}}\right]$

Where $v_y$ and $v_{0y}$ are components of velocities after and before collision along y-axis.

Hence, $e=-\left[\frac{v_y}{-v_0\text{cos}\theta }\right]\Rightarrow v_y=e{v}_0\text{cos}\theta$

Hence, the required height $=\frac{v_y^2}{2g}=\frac{e^2{v}_0^2{\text{cos}}^2\theta }{2g}$