 # Need help with this quadratic projectile motion problem

A golfer is standing on a tee block starting to drive and looking down on the fairway below (h = 0). The tee block is on a hill 10 yards above the fairway. The ball hit its max height of 60 yards, 125 yards away from the tee block. What would be the relation for this equation and how far would the ball have gone before it landed on the fairway?

The key, I think, to this problem, is understanding two very important facts:
(1) All motion under gravity is in the shape of a parabola.
(2) The max height of the golfball is at exactly the vertex of that parabola.

Using these two pieces of information, we can get started on the problem.

First, we need to situate the x-y axes. Since the golfer is 10 yards above the fairway, we’ll place the golfer at the point (0,10) on our graph. This means that the ball reaches its height at the point (125,60).

So (125, 60) is the vertex of our golf-ball parabola. So we can put this information in the equation y=a(x-h)^2 + k, which is a general parabola in vertex form. Since (h,k) is the vertex, just substitute to get: y = a(x-125)^2 + 60.

Alright! We’re almost done finding the relation for this equation. We just need to find a. To do this we just need to substitute any other point on the parabola and then isolate and solve for a. One other point that we have is (0,10), the initial position of the golfer. So make the substitution and solve:

10 = a(0-125)^2 + 60
15625a = 10-60 = -50
a = -\frac{50}{15625}
a = -0.0032.

FINAL EQUATION: So we have the final form of our equation: y = (-0.0032)(x-125)^2 + 60. This is what the graph looks like: HOW FAR DID THE BALL TRAVEL?
So, how far would the ball have gone before it landed on the fairway? We can see pretty clearly on the graph, but let’s figure it out algebraically. Just set y=0 and solve for x:

0 = (-0.0032)(x-125)^2 + 60
(x-125)^2 = \frac{60}{0.0032} = 18750
\sqrt{(x-125)^2} = \sqrt{18750}
x-125 \approx 136.9
x \approx 136.9+ 125 = 261.9

And then you’re done! Does all that make sense?

Yes, thank you. I just had a question that if 280 yards away their was a small creek running across the fairway, does the golfer need to worry about losing their ball? I tried to figure it out but I don’t know how.

Well, I’m not exactly sure what kind of answer that kind of question wants, but here’s my guess.
Since the ball lands at 261.9 yards and the creek is at 280 yards, it stands to reason that the golfer should be worried, since the ball could roll 18 more yards and fall into the creek. The golfer would only not have to worry if the ball landed farther than 280 yards, which it didn’t.

I hope you agree with my logic!

Yes I do, thank you so much and probably talk to you tomorrow for more help!

I’ll be here waiting! I appreciate your questions

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