The club was in contact with a ball, initially at rest, for about 0.0052 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 2.4 ✕ 102 ft/s, find the average force (in kN) exerted on the ball by the club.

2538.46N

Explanation:

Parameters given:

Initial speed, u = 0 m/s

Final velocity, v = 240 m/s

Time, t = 0.0052 s

Mass of club, m = 55g = 0.055 kg

First we find the acceleration:

a = (v - u)/t

a = (240 - 0)/0.0052

a = 46153.85 m/s²

Average force can then be found:

F = m*a

F = 0.055 * 46153.85

F = 2538.46 N

okay i have so many questions. but the big one here is: um wt..f is this mess? ?

the answer is a)

mark brainliest