A tennis ball of mass 0.0609 kg is served. It strikes the ground with a velocity of 52.5 m/s (117 mi/h) at an angle of 22.0° below the horizontal. Just after the bounce it is moving at 50.5 m/s at an angle of 17.7° above the horizontal. If the interaction with the ground lasts 0.0565 s, what average force did the ground exert on the ball?
To determine the average force exerted by the ground on the ball, we can use the impulse-momentum principle. The impulse-momentum principle states that the change in momentum of an object is equal to the force applied to the object multiplied by the time over which the force is applied.
We can break down the given information as follows:
- Initial velocity before the bounce: 52.5 m/s, at an angle of 22.0° below the horizontal.
- Final velocity just after the bounce: 50.5 m/s, at an angle of 17.7° above the horizontal.
- Mass of the tennis ball: 0.0609 kg.
- Interaction time with the ground: 0.0565 s.
First, let's calculate the initial momentum (pi) and final momentum (pf) of the tennis ball using the given velocities and the mass of the ball.
Initial momentum (pi) = mass × initial velocity
pi = 0.0609 kg × 52.5 m/s
Final momentum (pf) = mass × final velocity
pf = 0.0609 kg × 50.5 m/s
Next, we need to calculate the change in momentum (Δp) of the tennis ball.
Change in momentum (Δp) = pf - pi
Now that we have the change in momentum, we can calculate the average force (F) applied by the ground.
Average force (F) = Δp / interaction time
F = (pf - pi) / 0.0565 s
By substituting the values of pi, pf, and the interaction time, we can calculate the average force exerted by the ground on the tennis ball.