Final answer:
To calculate the time for which the ball remains in contact with the bat, the change in momentum must be determined by calculating the velocities before and after the hit. Using the impulse-momentum theorem and the average force applied, the contact time is found to be 1/80th of a second.
Step-by-step explanation:
The question involves applying the concept of impulse and the change in momentum when a ball is hit by a bat. According to the principle of impulse and momentum, the change in momentum of an object is equal to the impulse applied to it. The change in momentum (Δp) can be calculated by the difference in the final and initial momentum. Impulse (J) on the other hand, is the product of the average force (F) exerted on the object and the duration of time (t) for which the force acts.
To find the time for which the ball remains in contact with the bat, we first need to calculate the velocity the ball would have just before being hit by the bat. This is given by the equation for the velocity of a falling object: v = √(2gh), where g is the acceleration due to gravity and h is the height. For the ball dropped from 20m, the velocity at impact would be v = √(2 x 10 m/s² x 20 m) = √(400 m²/s²) = 20 m/s downwards.
Next, we apply the conservation of energy principle to find the velocity the ball must have after being hit to reach a height of 45m. Using the equation for potential energy (PE = mgh) and setting it equal to the kinetic energy (KE = 1/2 mv²) needed to reach that height, we get v = √(2gh) = √(2 x 10 m/s² x 45 m) = √(900 m²/s²) = 30 m/s upwards.
Considering the change in velocity (from 20 m/s downwards to 30 m/s upwards), the total change in velocity Δv is 50 m/s. Since the mass m of the ball is 50g (0.05 kg), the change in momentum Δp = m x Δv = 0.05 kg x 50 m/s = 2.5 kg m/s. By setting the impulse J equal to the change in momentum Δp, and knowing the average force F = 200N, we can solve for the contact time t: J = F x t = Δp, so t = Δp / F = 2.5 kg m/s / 200 N = 0.0125 s, which is 1/80th of a second.
Hence, the correct option is (C) 1/80th of a second.