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2.Motion in Straight Line
medium
A tennis ball is dropped on to the floor from a height of $9.8\,m$. It rebounds to a height $5.0\,m$. Ball comes in contact with the floor for $0.2\,s$. The average acceleration during contact is $..........ms ^{-2}$. [Given $g =10\,ms ^{-2}$ ]
A
$120$
B
$121$
C
$122$
D
$189$
(JEE MAIN-2023)
Solution
$v _{ i }=\sqrt{2 gh _{ i }}$
$=\sqrt{2 \times 10 \times 9.8} \downarrow$
$=14 m / s \downarrow$
$v _{ f }=\sqrt{2 gh _{ f }}$
$=\sqrt{2 \times 10 \times 5} \uparrow$
$= 1 0 ~ m / s \uparrow$
$\left|\overrightarrow{ a }_{\text {avg }}\right|=\left|\frac{\Delta \overrightarrow{ v }}{\Delta t }\right|=\frac{24}{0.2}=120\,m / s ^2$
Standard 11
Physics
