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2.Motion in Straight Line
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A ball is dropped on the floor from a height of $10 \,m$. It rebounds to a height of $2.5 \,m$. If the ball is in contact with the floor for $0.01 \,sec$, the average acceleration during contact is
A
$2100\,m/{\sec ^2}$ downwards
B
$2100\,m/{\sec ^2}$ upwards
C
$1400\,m/{\sec ^2}$
D
$700\,m/{\sec ^2}$
Solution
(b) Velocity at the time of striking the floor,
$u = \sqrt {2g{h_1}} = \sqrt {2 \times 9.8 \times 10} = 14\,m/s$
Velocity with which it rebounds.
$v = \sqrt {2g{h_2}} = \sqrt {2 \times 9.8 \times 2.5} = 7\;m/s$
$\therefore $ Change in velocity $\Delta v = 7 – ( – 14) = 21\,m/s$
$\therefore $ Acceleration $ = \frac{{\Delta v}}{{\Delta t}} = \frac{{21}}{{0.01}} = 2100\;m/{s^2}$ (upwards)
Standard 11
Physics
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