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A tennis ball with a speed of 27.7 m/s is moving perpendicular to a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 19.7224 m/s. If the ball is in contact with the wall for 0.0146 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/s 2 .

User Whhone
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1 Answer

4 votes

Answer:

a = 3248 m/s^2

Step-by-step explanation:

Given:

vi = 27.7 m/s

vf = -19.7224 m/s ***note the negative direction

t = 0.0146

a = ?

acceleration = change in speed / change in time

a = (27.7 - [-19.7224]) / 0.0146

a = 47.4224 / 0.0146

a = 3248 m/s^2

User Mludd
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