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A tennis ball with a speed of 22.2 m/s is moving perpendicular to a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 14.9184 m/s. If the ball is in contact with the wall for 0.00965 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

Respuesta :

Answer:

[tex]a=-3846.46\ m/s^2[/tex]

Explanation:

Given that,

Initial velocity of the ball, u = 22.2 m/s

Final velocity of the ball, v = -14.9184 m/s (minus sign shows opposite direction of motion)

Time for contact, t = 0.00965 s

We need to find the average acceleration of the ball when it is in contact with the ball. The acceleration of an object is equal change in velocity per unit time. So,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{-14.9184 -22.2 }{0.00965 }\\\\a=-3846.46\ m/s^2[/tex]

So, the acceleration of the ball is [tex]3846.46\ m/s^2[/tex] towards the opposite direction of motion.

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