Question

To test the quality of a tennis ball, you drop it onto the floor from a height of $4.00m$. It rebounds to a height of $2.00m$. If the ball is in contact with the floor for $12.0ms$, what is its average acceleration during that contact? Take $g=10m/s^2.$

• A. $1.26\times {10}^{3}m{s}^{-2}$ (upward)
• B. $1.26\times {10}^{3}m{s}^{-2}$ (downward)
• C. $2.52\times {10}^{3}m{s}^{-2}$ (upward)
• D. $2.52\times {10}^{3}m{s}^{-2}$ (downward)

Answer 1

The correct option is A $1.26\times {10}^{3}m{s}^{-2}$ (upward)

Using, $v^2=u^2+2as$
Velocity of ball just before hitting the floor
$v_1^2=2\times 10\times 4$
$or{v}_1=\sqrt{80}m/sdownwards$
Velocity of ball after hitting the floor
$0=v_2^2-2\times 10\times 2$
$or{v}_2=\sqrt{40}m/supwards$

Using $v=u+at$ for $12ms$ time (taking upwards direction as positive)
$\sqrt{40}=-\sqrt{80}+a\times 0.012$
$\Rightarrow a=\frac{\sqrt{40}+\sqrt{80}}{0.012}$
$\Rightarrow a=1272.4m/s^2\cong 1.26\times {10}^{3}m/s^{2}upwards$
Thus correct option is (a)