Respuesta :
Answer:Racquet force is twice of Player force
Explanation:
Given
ball arrives at a speed of [tex]u=-40\ m/s[/tex]
ball returned with speed of [tex]v=40\ m/s [/tex]
average Force imparted by racquet on the ball is given by
[tex]F_{racquet}=\frac{m(v-u)}{\Delta t}[/tex]
where [tex]m=mass\ of\ ball[/tex]
[tex]\Delta t=[/tex]time of contact of ball with racquet
[tex]F_{racquet}=\frac{m(40-(-40))}{\Delta t}[/tex]
[tex]F_{racquet}=\frac{80m}{\Delta t}-----1[/tex]
When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet
[tex]F_{player}=\frac{m(0-40)}{\Delta t}[/tex]
[tex]F_{player}=\frac{-40m}{\Delta t}-----2[/tex]
From 1 and 2 we get
[tex]F_{racquet}=-2F_{player}[/tex]
Hence the magnitude of Force by racquet is twice the Force by player
The comparison of both forces for the first and second player is;
The magnitude of force exerted by the first player's racquet on the ball is twice the magnitude of the force exerted by the second player's hand on the ball.
We are given;
Initial velocity of tennis ball; u = 40 m/s
Final velocity of tennis ball on return; v = -40 m/s (negative because it is equal and opposite to the initial velocity)
Let mass of the player be m. Thus;
Force by 1st player's racket on ball; F_racquet = m(v - u)/t
F_racquet = m(-40 - 40)/t
F_racquet = -80m/t ----(eq 1)
The second player now catches the ball in her hands. This means final velocity for second player is v = 0 m/s.
Thus, force exerted by the second player's hand on the ball is;
F_hand = m(0 - 40)/t
F_hand = -40m/t ----(eq 2)
In conclusion, can see that F_racquet is twice F_hand.
Read more at; https://brainly.com/question/14818845
