First slide

Average Acceleration

Question

A ball dropped on to the floor from a height of 10 m rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.02s, its average acceleration during contact is

Moderate

A

2100ms^-2
B

1050 ms^-2
C

4200 ms^-2
D

9.8 ms^-2

Solution

$\large a = \frac{{v - u}}{t} = \frac{{\sqrt {2g{h_2}} + \sqrt {2g{h_1}} }}{t} = \frac{{\sqrt 2 g(\sqrt {{h_2}} + \sqrt {{h_1}} )}}{t}$
$\large = \sqrt {2 \times 9.8} \frac{{(\sqrt {2.5} + \sqrt {10} )}}{{0.02}} = 1050{\kern 1pt} {\kern 1pt} m/{s^2}$

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