Ball moving up an incline and coming down.

  • Thread starter bolas
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  • #1
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"a ball is rolled up an incline (at an angle x with the horizontal). For the angle x and the ball involved, the acceleration of the ball is 0.25g and directed down the incline. If the ball is released with speed v, determine the distance it moved up the incline before reversing its direction as a term of all other variables. What is the minimum initial velocity?"

Anyone can give me a hint on how to approach this problem? :(
 
Last edited:

Answers and Replies

  • #2
radou
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Well, try to present us some of your work, there must be some thoughts on how to solve the problem.
 
  • #3
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This is what I have so far..

Using equation of motion..

Vf^2 = Vi^2 + 2*a*x

a = 0.25g = 0.25*(-9.8 m/s^2) = -2.45 m/s^2

Vf = 0 m/s
Vi = ?
x = s = ?

0 = Vi^2 + 2(-2.45 m/s^2)x
0 = Vi^2 - 4.9 m/s^2 * x

-Vi^2 = -4.9 m/s^2 * x

Vi^2 = 4.9 m/s^2 *x

x = vi^2/4.9 m/s^2
 
  • #4
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Gave up? :(
 
  • #5
radou
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Gave up? :(
No need to give up, since the answer seems correct. :smile:
 
  • #6
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For this problem based on the given information that th acceleration is 0.25g down the incline plane, can I assume then that the angle is...

0.25g = gsin(theta)
theta = 14.47 deg ?
 
  • #7
radou
Homework Helper
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For this problem based on the given information that th acceleration is 0.25g down the incline plane, can I assume then that the angle is...

0.25g = gsin(theta)
theta = 14.47 deg ?
You have nothing to assume, since the accelearion is 0.25g, which must equal g sin(theta), so theta ≈ 14.47 deg.
 

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