To determine the force exerted by the floor on the south end of the court on the tennis ball, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Mass of the tennis ball (m) = 0.1 kg
Acceleration of the tennis ball (a) = 10 m/s^2 (assuming the given value is in meters per second squared)
First, let's convert the velocity of the racket from km/h to m/s:
Velocity of the racket = 65 km/h
= 65 × (1000/3600) m/s
= 18.06 m/s
Since the ball is hit by the racket, it inherits the velocity of the racket. Therefore, the initial velocity of the ball is also 18.06 m/s.
Using the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as the ball hits the floor)
u = initial velocity (18.06 m/s)
a = acceleration (10 m/s^2)
s = displacement (let's assume the displacement to be the length of the court)
Let's solve for the displacement (s):
0^2 = 18.06^2 + 2(10)s
0 = 326.7636 + 20s
20s = -326.7636
s = -16.33818 m
Since the displacement value is negative, it means the ball moves in the opposite direction (from north to south).
Now, let's calculate the force exerted by the floor on the ball:
F = m * a
F = 0.1 kg * 10 m/s^2
F = 1 N
Therefore, the floor on the south end of the court exerts a force of 1 Newton (N) on the tennis ball.
a tennis ball player hits a 0.1 kilogram tennis ball with her racket from the north end of a court. her racket was traveling at 65 kilometers per hour. the ball accelerated at a rate of 10 meters. the ball hits the floor on the south end of the tennis court. how much force did the floor on the south end of the court exert on the ball
1 answer
1 answer