Golf ball at 17° to reach 250 m green with same height

[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.

Homework Equations

trig equations
v2^2=v1^2+2ad
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.

 

[Dr Transport]

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

 

[haruspex]

dy (tan17*250)

Wouldn't that be the height by which the ball would fly over the green if there were no gravity?

 

[haruspex]

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.

 

[gungo]

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

 

[haruspex]

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.

 

[gungo]

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?

 

[haruspex]

But then we have nothing to input into the equations

Sure you do. Try it.

 

[gungo]

Sure you do. Try it.

dy:0 dx:250 v2^2=v1^2+2ad ?=?+2(-9.8)(0)
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

 

[gneill]

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.

 

[gungo]

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

 

[haruspex]

dy:0 dx:250 v2^2=v1^2+2ad ?=?+2(-9.8)(0)
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.