Q 11-02-072JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
A tennis ball is dropped on to the floor from a height of $9.8\ \text{m}$. It rebounds to a height $5.0\ \text{m}$. Ball comes in contact with the floor for $0.2\ \text{s}$. The average acceleration during contact is ______ $\text{m s}^{-2}$. [Given $g = 10\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 120
Speed just before impact (downward): $v_1=\sqrt{2gh_1}=\sqrt{2\times10\times9.8}=14\ \text{m s}^{-1}$.
Speed just after impact (upward): $v_2=\sqrt{2gh_2}=\sqrt{2\times10\times5}=10\ \text{m s}^{-1}$.
Taking upward as positive, the change in velocity is $10-(-14)=24\ \text{m s}^{-1}$, so
$$a_{avg}=\frac{\Delta v}{\Delta t}=\frac{24}{0.2}=120\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics