What is the Acceleration of a Tennis Ball After Being Served?

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To determine the acceleration of a tennis ball served at 49 m/s over a distance of 45 cm, one can use kinematic equations. The relevant equations are v = at and s = (1/2)at², where v is final velocity, a is acceleration, and s is distance. By substituting the known values into these equations, the acceleration can be calculated without needing time. Once acceleration is found, the net force can be calculated using Fnet = ma. The discussion emphasizes the importance of using kinematic formulas correctly to solve for acceleration and net force.
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When a 58 g tennis ball is served, it accelerates from rest to a constant speed of 49 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 45 cm. What is the magnitude of the net force acting on the ball?

I tried using Fnet=ma, but acceleration isn't given. So I tried to find acceleration by dividing the velocity by time, but time isn't given. Then I tried to use a kinematic formula without time in it to solve for acceleration. When I got my "answer", it wasn't right. What in the world am I doing wrong?
 
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How much work has the force done on the ball?
 
For any acceleration, starting from rest, v= at and s= (1/2)at2.

Here you are told that v= 49 m/s and s= 45 m/s2. You can solve those two equations for a and t. (Of course, you only need a.)
 

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