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**1. I think the most place I'm questioning my answer is the force part of it.**

A golf club contacts a golf ball for 0.15 s at an angle to the horizontal and the ball subsequently strikes the ground a distance down the fairway. You choose an initial speed (between 25 m/s and 35 m/s) and the angle of projection (Between 15o and 22o) to calculate

the time between the ball leaving the club and when it first touches the ground,

the range of the ball down the fairway , and

the average force exerted by the club on the ball.

(assume the fairway is level horizontally. Mass of a golf ball is 0.046 kg)

vh = 28.19 m/s cos 20 = vh/30

vv = 10.26 m/s sin 20 = vv/30

t = (0.0 m/s - 10.26 m/s) / -9.8

t = -10.26 / -9.8

t= 1.047 s

t= 2 (1.047)

t= 2.094 seconds for the ball to touch the ground

dx = (28.19 m/s) (2.094 s)

dx= 59.03 meteres

f=.046(-9.8)

f=-.45N

Homework Statement

A golf club contacts a golf ball for 0.15 s at an angle to the horizontal and the ball subsequently strikes the ground a distance down the fairway. You choose an initial speed (between 25 m/s and 35 m/s) and the angle of projection (Between 15o and 22o) to calculate

the time between the ball leaving the club and when it first touches the ground,

the range of the ball down the fairway , and

the average force exerted by the club on the ball.

(assume the fairway is level horizontally. Mass of a golf ball is 0.046 kg)

## Homework Equations

vh = 28.19 m/s cos 20 = vh/30

vv = 10.26 m/s sin 20 = vv/30

t = (0.0 m/s - 10.26 m/s) / -9.8

t = -10.26 / -9.8

t= 1.047 s

t= 2 (1.047)

t= 2.094 seconds for the ball to touch the ground

dx = (28.19 m/s) (2.094 s)

dx= 59.03 meteres

f=.046(-9.8)

f=-.45N

## The Attempt at a Solution

Homework Statement

The ball will take 2.094 seconds to touch the ball 59.03 meteres down the fairway forced at -.45N.