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andreev551 [17]
3 years ago
14

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

yer, realizing that the ball is out of bounds, catches it in her hand. Assuming the time interval of contact is the same in both cases, compare the magnitude of the force exerted by the first player's racquet on the ball with the magnitude of the force exerted by the second player's hand on the ball
Physics
1 answer:
Butoxors [25]3 years ago
5 0

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of u=-40\ m/s

ball returned with speed of v=40\ m/s

average Force imparted by racquet on the ball is given by

F_{racquet}=\frac{m(v-u)}{\Delta t}

where m=mass\ of\ ball

\Delta t=time of contact of ball with racquet

F_{racquet}=\frac{m(40-(-40))}{\Delta t}

F_{racquet}=\frac{80m}{\Delta t}-----1

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

F_{player}=\frac{m(0-40)}{\Delta t}

F_{player}=\frac{-40m}{\Delta t}-----2

From 1 and 2 we get

F_{racquet}=-2F_{player}

Hence the magnitude of Force by racquet is twice the Force by player

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Answer:

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By solving the above equation for t we have:

t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s  

Hence, the ball will stay 0.55 seconds in the air.

                             

B) We can find the distance traveled by the ball as follows:

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

Where:  

a: is the acceleration in the horizontal direction = 0  

x_{f} is the final position =?  

x_{0} is the initial position = 0      

v_{0x} is the component of the initial speed in the horizontal direction = 45 m/s                                                                                            

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

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