A tennis player smashes a serve so that the racquet is in contact with the ball for 0.055 s, giving it an impulse of 2.5 N ⋅ s. What average force was applied during this time? Assume that the vertical motion of the ball can be ignored. If the ball's mass is 0.060 kg, what will be the ball's horizontal velocity?

force*time=impulse solve for force.

then finally, impluse=mass*velocity

To find the average force applied during the contact time, we can use the formula:

Average force = impulse / contact time

Given:
Impulse (J) = 2.5 N ⋅ s
Contact time (Δt) = 0.055 s

Substituting the values into the formula:

Average force = 2.5 N ⋅ s / 0.055 s
Average force = 45.45 N

Therefore, the average force applied during the contact time is 45.45 N.

To find the ball's horizontal velocity, we can use the formula for velocity:

Velocity (v) = impulse / mass

Given:
Impulse (J) = 2.5 N ⋅ s
Mass (m) = 0.060 kg

Substituting the values into the formula:

Velocity = 2.5 N ⋅ s / 0.060 kg
Velocity = 41.67 m/s

Therefore, the ball's horizontal velocity will be 41.67 m/s.