Calculate the allowable range of initial velocities

  • Thread starter Thread starter electro05
  • Start date Start date
  • Tags Tags
    Initial Range
AI Thread Summary
The discussion revolves around calculating the initial velocities required for a golfer to stop a ball within a 1.0m range of a cup, considering uphill and downhill conditions. The uphill scenario has a constant deceleration of 2.0m/s², while the downhill has a deceleration of 3.0m/s². Participants suggest using the kinematic equation to find the initial velocities needed to stop the ball 1 meter short and 1 meter long from the cup. The calculations should be performed separately for both uphill and downhill scenarios to establish the complete range of allowable velocities. This approach ensures that the golfer can accurately control the ball's stopping position relative to the cup.
electro05
Messages
3
Reaction score
0
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:
Physics news on Phys.org
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.
 
thanks much!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top