Question

Q.17. A ball is dropped from a height of 7.2 m. It bounces back to 3.2 m after striking the floor, the ball remains in contact with the floor for 20 ms. Given that g = 10 ms-2, the average acceleration of the ball during the contact is :
(A) 100 $m{s}^{-2}$
(B) 200 $m{s}^{-2}$
(C) 600 $m{s}^{-2}$
(D) 1000 $m{s}^{-2}$

Answer 1

$$\begin{gathered} Dearstudent, \\ {a}_{avg}=\frac{v-u}{t} \\ {u}^2=0^2+2\times 10\times 7.2 \\ \Rightarrow u=12m/s \\ Forvwehave, \\ {0}^2=v^2-2\times 10\times 3.2 \\ \Rightarrow v=8m/s \\ {a}_{avg}=\frac{8-(-12)}{20\times {10}^{-3}}=1000m/s^2 \\ Regards \end{gathered}$$