# Projectile Motion: golf ball velocity to reach the green

• gungo
In summary: Since you have the launch angle, the vertical requirements (d=0) give you a relationship between the initial speed and the flight time.The horizontal requirements give you another relationship between them.Solve the simultaneous equations.The projectile will fly over the green if it has an initial speed of 66m/s and the launch angle is 17 degrees.
gungo

## Homework Statement

A golfer strikes a golf ball at an angle of 17 degrees above the horizontal. With what velocity must the ball be hit to reach the green which is a horizontal distance of 250 m from the golfer? It is at the same height as the tee.

trig equations
d=v1(t)+1/2a(t)2
d=(v1+v2/2)t

## The Attempt at a Solution

Well I used trigonometry to find the dy (tan17*250) and I believe that's the maximum height...meaning v2y at that time is 0
So I have dy=76.43 m
a=-9.8m/s^2
v2y=0
I used v2^2=v1^2+2ad to find the initial velocity in the y direction and got 38.7
I plugged that into the equation d=(v1+v2/2)t to find time and got 1.975 seconds
For the x direction I know dx=250, a=0, and t=1.975 seconds
So I used d=v1(t)+1/2a(t)2 to find v1 but the a is 0 so the equation just ends up being d=v1(t) and I got the initial velocity in the x direction to be 126.59
I used pythagorean theorem to find the total velocity to be 132.34 m/s 17 degrees above the horizontal...but the answer is 66m/s 17 degrees above the horizontal.

use the range equation $R = \frac{v_{0} ^{2}sin(2\theta)}{g}$

gungo said:
dy (tan17*250)
Wouldn't that be the height by which the ball would fly over the green if there were no gravity?

Dr Transport said:
use the range equation $R = \frac{v_{0} ^{2}sin(2\theta)}{g}$
That would certainly be simpler, but I prefer first to help the student fix the errors in the chosen method.

haruspex said:
Wouldn't that be the height by which the ball would fly over the green if there were no gravity?
Sorry I don't understand

gungo said:
Sorry I don't understand
You have not posted a diagram, but I hope you drew one.
The launch angle is 17 degrees and the length of the hole 250m. To reach a height of 250 tan 17 it would have to keep going in a straight line (so no gravity) for the entire distance.

haruspex said:
You have not posted a diagram, but I hope you drew one.
The launch angle is 17 degrees and the length of the hole 250m. To reach a height of 250 tan 17 it would have to keep going in a straight line (so no gravity) for the entire distance.
So dy isn't 250 tan 17...it's 0 because it lands on the ground. But then we have nothing to input into the equations?

gungo said:
But then we have nothing to input into the equations
Sure you do. Try it.

haruspex said:
Sure you do. Try it.
t:? t:? v2=v1+at ?=?+-9.8?
a:-9.8 a:0 d=v1(t)+1/2(a)(t)^2 0=?(?)+1/2(-9.8)(?)^2
vy1:? vx1:? d=v2(t)-1/2(a)(t)^2 0=?(?)-1/2(-9.8)(?)^2
vy2:? vx2:? d=(v1+v2/2)(t) 0= (?+?/2)(?)
I don't know what missing information I'm able to get from the two numbers given in the original question? I can't think of anything

Moderator note: Thread title changed to make it descriptive of the problem. Thread titles should not be overly generic! "Projectile motion" was far too generic.

gneill said:
Moderator note: Thread title changed to make it descriptive of the problem. Thread titles should not be overly generic! "Projectile motion" was far too generic.
whoops sorry

gungo said:
t:? t:? v2=v1+at ?=?+-9.8?
a:-9.8 a:0 d=v1(t)+1/2(a)(t)^2 0=?(?)+1/2(-9.8)(?)^2
vy1:? vx1:? d=v2(t)-1/2(a)(t)^2 0=?(?)-1/2(-9.8)(?)^2
vy2:? vx2:? d=(v1+v2/2)(t) 0= (?+?/2)(?)
I don't know what missing information I'm able to get from the two numbers given in the original question? I can't think of anything
Since you have the launch angle, the vertical requirements (d=0) give you a relationship between the initial speed and the flight time.
The horizontal requirements give you another relationship between them.
Solve the simultaneous equations.

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air under the force of gravity. It follows a curved path known as a parabola.

## 2. How is golf ball velocity calculated?

Golf ball velocity is calculated by dividing the distance the ball travels by the time it takes to travel that distance. This can be done using a radar or launch monitor.

## 3. What factors affect the velocity of a golf ball?

The velocity of a golf ball can be affected by factors such as the force of the swing, the angle of the clubface, the loft of the club, and the air resistance or drag.

## 4. How can I determine the velocity needed to reach the green?

The velocity needed to reach the green can be determined by considering the distance to the green, the angle of the shot, and any obstacles or hazards in the way. This can also be calculated using physics equations and taking into account the launch angle and initial velocity.

## 5. What is the ideal velocity for a golf ball to reach the green?

The ideal velocity for a golf ball to reach the green depends on the distance to the green, the type of club being used, and the golfer's skill level. Generally, a lower velocity is needed for shorter shots, while longer shots may require a higher velocity to overcome the effects of air resistance and reach the green.

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