A ball hits a wall horizontally at $6.0\ ms^{-1}$. It rebounds horizontally at $4.4\ ms^{-1}$. The ball is in contact with the wall for 0.040 s. What is the acceleration of the ball?

Academic Physics NCERT Class 9

Here initial velocity $u=6.0\ ms^{-1}$

Final velocity in opposite direction $v=-4.4\ ms^{-1}$

Time $t=0.040\ s$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{-4.4-6.0}{0.040}$

$=\frac{-10.4}{0.040}$

$=-260\ ms^{-2}$

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Updated on: 10-Oct-2022

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