Range formula on planets

  • #1
249
1

Homework Statement


On Earth you are able to consistently hit a golf ball 156 meter when you use your 5 iron which launches the ball at a 23 angle with respect to the ground. If you were to perform this same shot on the surface of an unknown planet called x, and be able to hit the ball 200 m, what will be the acceleration due to gravity on the surface of the planet x?

My confusion is that I have somehow ended up with three range formulas and I don't know what the difference between them is:

Homework Equations


R1 = Vo2 sin(2θ)/g

R2 = Vo2 sin(2θ)/2g

I also have my textbook saying:
h = Vo2 sin2(2θ)/2g

Vo = initial velocity

The Attempt at a Solution



156 meters = V2osin(46)/(2*9.8)

Vo = 65.1964 m/s

Then solving for g on planet x I get the answer as 7.644 which is the correct answer.

What confuses me is the difference between the formulas. Are two of them wrong or are they applicable in different circumstances?
 
Last edited:

Answers and Replies

  • #3
249
1
Yes, but using that formula does not give me the correct answer.
 
  • #4
294
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I'm confused, if you are able to hit a ball the same distance with the same stroke on two different planets, wouldn't that mean that gravity was the same on each?
 
  • #5
249
1
That's because I forgot part of the question. On planet X when you hit the ball it goes 200 m.

Sorry, 'll go back and edit the question.
 
  • #6
294
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Ah okay :)

You said that you get the correct answer when you use the formula with a ##2g## in the denominator instead of just ##g##, and when you use the ##g## formula you get it wrong.

But when you use the formula with just a ##g##, did you recalculate your ##V_o##?
 
  • #7
249
1
I ran the calculations again and you're right. I must've confused myself and used the wrong numbers somewhere. Thanks :)

So I get

R=Vo^2 sin(2a)/g

156 m= Vo^2 sin(46)/9.8

Vo = 46.1 m/s

200 m = 46.1^2sin(46)/g

g = 7.64 m/s^2
 

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