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A tennis ball with a speed of 24.4 m/s is moving perpendicular to a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 19.7884 m/s. If the ball is in contact with the wall for 0.0145 s, what is the average acceleration of the ball while it is in contact with the wall? Take "toward the wall" to be the positive direction. Answer in units of m/s² .

Answer :

Answer:

The acceleration of the ball is [tex]3047.4758m/sec^2[/tex]

Explanation:

We have given that ball strikes the wall at a speed of 24.4 m/sec

And after striking its goes in opposite direction with a speed of 19.7884 m/sec

We have given that speed towards the wall is positive

So u = 24.4 m/sec

And v = -19.7884 m/sec

Time is given as t = 0.0145 sec

We know that acceleration is given by [tex]a=\frac{v-u}{t}=\frac{24.4-(-19.7884)}{0.0145}=3047.4758m/sec^2[/tex]

So the acceleration of the ball is [tex]3047.4758m/sec^2[/tex]

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