A golf club striking a golf ball imparts a force of 2300 N over a time of 4.36×1

Question

A golf club striking a golf ball imparts a force of 2300 N over a time of 4.36×103 s at an angle of 35.0 above the horizontal. Answer the following questions concerning the motion of the golf ball if it has a mass of 0.150 kg

With what speed does the ball leave the club head after impact?

What maximum height does the golf ball achieve after impact?

What horizontal distance will the golf ball travel before striking the ground? Assume the ground is perfectly level and that air resistance has no effect on the flight of the ball.

Explanation / Answer

Given
F = 2300N
t = 4.36*10^-3 s
angle = 35 degrees
m = 0.150 kg

Initial velocity of ball (u) = momentum gained (kg.m/s) / mass (kg)
= impulse (N.s) / mass

u = (2300* 4.36*10^-3) / 0.15
u = 66.853 m/s

A) Equ. for max.height
h = (u.sin)^2 / 2g
h = ((66.853*sin(35))^2)/(2*9.8)
h=75.1 m

B) Equation for max.range
R= u².sin2 / g

R = 66.853*66.853*sin(2*35)/9.8
R = 428.55 m

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