Projectile Motion Question 1: Calculating Time, Range, and Force in Golf Swing

AI Thread Summary
The discussion focuses on calculating the time, range, and force involved in a golf swing impacting a golf ball. The calculations indicate that the ball will take approximately 2.094 seconds to hit the ground, traveling a distance of 59.03 meters. The average force exerted by the club on the ball is initially calculated as -0.45 N, but this value is questioned due to the assumption that the ball is accelerating at 9.8 m/s² when struck. Participants suggest reevaluating the force calculation by considering the contact time of 0.15 seconds and the change in velocity. Accurate force determination is emphasized as crucial for solving the problem correctly.
billybobay
Messages
13
Reaction score
0
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)


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.
 
Physics news on Phys.org
vh = 28.19 m/s cos 20 = vh/30
vv = 10.26 m/s sin 20 = vv/30
Okay, looks like you chose the initial angle to be 20 degrees and the initial speed to be 30. Then your vh and vy are correct.
The time to touch down and range look good.

The F = ma = .046(-9.8) can't be right. The ball is not accelerating at 9.8 when hit by the club. Your are given that "club contacts a golf ball for 0.15 s". Perhaps you can find some way to calculate the force from that and its change in velocity.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top