A 0.061 kg golf ball is struck by a golf club with a force of 299.4 N. The change in velocity of the ball is 54.3 m/s. How much time is the ball in contact with the club?

The time the ball is in contact with the club is approximately 0.011063 seconds

Explanation:

The question is with regards to Newton's second law of motion which states that a force is equal to the rate of change of momentum produced

The given parameters are;

The mass of the ball, m = 0.061 kg

The force with which the ball is struck with the golf club, F = 299.4 N

The change in velocity of the ball, Δv = 54.3 m/s

By Newton's second law, we have;

Where;

Δt = The time it takes the momentum of the object to change = The time the ball is in contact with the club

Substituting the known values, we get;

The time the ball is in contact with the club = Δt ≈ 0.011063 seconds.

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.