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Calculate the allowable range of initial velocities

  • Thread starter electro05
  • Start date
3
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Hi all. I'm stuck on a physics problem obviously...So the original problem basically is a golfer is trying to put a ball 1.0m long or short of the cup. From uphill it is more difficult than downhill-explain why. Assume that the ball decelerates constantly at 2.0m/s^2 going downhill, and constantly at 3.0 m/s^2 going uphill. The uphill and downhill lie are both 7.0m from the cup. I'm supposed to calculate the allowable range of initial velocities that can be imparted on the ball so that it stops in the 1.0m long or short range from the cup.

I wrote down my given for this: (hope i got it right)
a(up)=2.0m/s^2
a(down)=-3.0m/s^2
V(init)=?
V(final)=0m/s
x(down)=-7.0m
x(up)=7.0m (should the distance be different since the golfer can hit it anywhere between 6m-8m?)

I tried using the V(init)=sqrt[V(final)-2a(x-x(init)], but i don't think i'm doing it right. Can someone give me some pointers or explain this problem a bit more? Thanks for the help!
 
Last edited:

Answers and Replies

107
0
You have the correct equation.
Calculate the velocity to get it to stop 1 meter in front from uphill. Then calculate the velocity for 1 m behind for uphill. Now you have your range of allowable velocities for hitting the ball from uphill. Repeat the same process for the downhill portion of the problem.
 
3
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thanks much!
 

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