Tennis player hits a 0.1 kilogram tennis ball with her racket from the north end of the court. Her racket was traveling at 65 kilograms per hour. The ball accelerated at a rate of 10 meters per second squared. The ball hits the floor on the south end of the tennis court. The floor sends the ball toward the tennis players opponent with the same acceleration. How much force did the floor on the south end of the court exert on the ball
First, we need to calculate the initial velocity of the ball when it leaves the racket. We can convert the racket's velocity from km/h to m/s:
65 km/h = 65 * 1000m / 3600s = 18.0556 m/s
Next, we can calculate the initial velocity of the ball when it leaves the racket:
v = u + at
v = 18.0556 m/s + 0 * 0.1 s
v = 18.0556 m/s
Now, we can calculate the force exerted by the floor on the ball when it hits the floor. We can use the equation:
F = ma
Where,
m = 0.1 kg (mass of the ball)
a = 10 m/s^2 (acceleration of the ball)
F = 0.1 kg * 10 m/s^2
F = 1 N
Therefore, the floor exerts a force of 1 Newton on the ball when it hits the floor on the south end of the court.