Heighth of a golf ball

Z

Zekette

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Joined
Feb 23, 2005
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A golf ball goes 40 yards in 1.31 seconds. Theda=60 degrees
How far it goes is x=rcos(theda)t [this is t times cos theda]
How high it is at "t" is y=rsin(theda)t-16(t squared)

I got that r=61.07, but am stuck after that. Thank you for your help![/img]
 
Hello, Zekette!

Yes, please give us the <u>original</u> wording of the problem.
There are serious gaps in this version.

A golf ball goes 40 yards in 1.31 seconds.
Does this mean that the ball hits the ground then ... or is it still in the air?

Theta=60 degrees . . . I assume this is the angle of elevation

How far it goes is: x = r(cosθ)t

How high it is: y = r(sinθ)t - 16t<sup>2</sup>

I got that r = 61.07, . . . not correct **
but am stuck after that. . . . Stuck on what? What was the question?
Click to expand...
** From the "16" in the equation, the distances are in <u>feet</u>.
. . . And 40 yards = 120 feet.
 
Sorry

Sorry for the lack of information. The question is how high does the golf ball go? As in, it's maximum heighth. R in feet would be 183.2? I don't know how to go any farther than that. THANK YOU!!
 
Taking what you wrote literally I get
40*3 = 120 = r*cos(60)*1.31
r = 183.2 ft/sec
y = r*sin(60)t-16t^2 =
183.2*.866*t-16t^2 =
158.66t-16t^2
The vertex is t = -b/2a =
-158.66/(2*(-16)) =
4.96 seconds
y = 158.66*4.96-16((4.96)^2 =
393 feet high.
Incidently it goes about 909 feet before it hits the ground. That's some drive. Must have ued a shotgun.
 
Um

I'm a little confused as to what you were doing. I think I have an idea of how to do it. This is what I did:
I took the derivative of the y= equation and got rcos(theda)-32t
I set this equal to 0 to find the extremum point, and got t=2.8625
I put it back into the equation so I had: y=183.2sin60(degrees)*2.8625-16*(2.8625^2)
y=323 ft
Is this correct?
 
As I said, I took you at your word.
A golf ball goes 40 yards in 1.31 seconds
Click to expand...
How far it goes is x=rcos(theda)t
Click to expand...
Putting those two statements together (Nothing is said about the ball hitting the ground) lets you solve for r which is the velocity of the ball.
That's where I got
40*3 = 120 = r*cos(60)*1.31
r = 183.2 ft/sec
Then I used r in the second equation to get
y = r*sin(60)t-16t^2 =
183.2*.866*t-16t^2 =
158.66t-16t^2
I used the vertex formula to find the t of the vertex.
The vertex is t = -b/2a =
-158.66/(2*(-16)) =
4.96 seconds
Then I solved the second equation for the height at that time.

As to your solution. You used theta as a variable. It is a constant (60 degrees) so the derivitive is
rsin(theta)-32t
If you use that you get the same (strange) answer as I did.
 
The ball hits the ground 40 yds away after 1.31 seconds.
How do you have a b and an a?
 
YOU KEEP CHANGING THE PROBLEM
Why won't you type what the book says?
Everyone who has tried to help you has asked you to do that and you keep ignoring us.
I would be fascinated to see how you solve
120 = r*cos(60)*(1.31)
and
0 = r*sin(60)*(1.31)-16*(1.31)^2
for the same value of r which would have to be true if if the ball hits the ground (height zero) at 40 yards after 1.31 seconds.

The a and b come from the standard quadratic equation.
y = ax² + bx + c
Haven't you met the quadratic solution?
x = (-b+sqrt(b²-4ac))/2a
I'm using the same a and b. In fact the vertex equation comes from that. Same -b and 2a.
------------------
Gene
 
There was no book, that was the information our teacher gave us. But, it's ok, i got his help on it. Thank you anyways!
 
I would like to know the answer too. Would you share your work?
---------------------
Gene
 
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