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Homework Help: Conservation of energy

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Drew kicks a ball (mass=0.250 kg) off his back porch at h= 1.50 m above the ground with an initial speed v= 17.0 m/s at some known angle. Find the initial kinetic energy of the ball. Find the initial potential energy of the ball. Find the kinetic energy and the speed v of the ball when it reaches the top of the ballistic arc, at h = 9.50 m. Find the speed of the ball v when it reaches the ground h=0.

    2. Relevant equations

    3. The attempt at a solution

    Initial kinetic energy:
    KE = 1/2 mv ^2
    KE = 1/2 (0.250) (17.0) = 36. 125 J

    Initial potential energy:
    PE = mgh
    PE= (0.250) (9.81) (1.50) = 3.679 J

    This is where I got stuck. If I do not know the speed of the ball when it reaches the top of the arc, how am I supposed to find the KE?
    I tried to solve for v by doing:
    1/2mv^2 + mgh = 1/2mv^2 + mgh

    v^2 = 2(9.81) (9.50)
    took the square root..and got v = 13.65 m/s
    Then i plugged this speed into the KE equation to solve for KE.
    Is this correct? It doesn't seem right to me. I remember hearing that at the top of an arc v would equal zero, but I don't know if that only applies with projectile motion.

    Help would be appreciated! Thanks
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 23, 2012 #2
    At the top of the arc Vertical component of velocity is zero , not the horizontal :D
  4. Feb 23, 2012 #3
    Thanks! And for the last part of the problem at h=0 would the final speed be the same as the initial speed?
  5. Feb 23, 2012 #4
    For the initial kinetic equation, do not forget to square your velocity.
  6. Feb 23, 2012 #5
    Oh and if I remember correctly, when the ball is at the top of the arc, its velocity is zero. The kinetic energy is transferred all into gravitational energy. Gravitational energy is a type of potential energy.

    But please don't take my word for it until someone confirms my statement.
  7. Feb 23, 2012 #6
    Chemicist: I thought it would equal zero too that is where I am still confused...
  8. Feb 23, 2012 #7
    Yes it would equal zero. Because in a system, no energy is created or destroyed. So the kinetic energy, or the energy of motion, is converted into pure gravitational energy (potential) at its peak height, which is the top of the arc.
  9. Feb 23, 2012 #8
    if the velocity is zero at top what makes the ball go forward in horizontal direction?
  10. Feb 23, 2012 #9
    So, if v=0 then ke would = 0 as well because the energy is converted into gravitational energy?
  11. Feb 23, 2012 #10
    I believe so, but again, don't turn this in until someone can confirm my statement. I don't want to be giving you wrong answers and if I am, I want someone to correct them.
  12. Feb 23, 2012 #11
    I don't believe it moves in a horizontal direction because the energy is converted into gravitational, which is the energy amassed by a height. So once it's at its peak, it has 100% gravitational energy and the force of gravity will push the ball downward.
  13. Feb 23, 2012 #12
    According to you it will push downward but ball moves forward as well as downward it follows a trajectory , the thing which you are telling happens when you throw a ball straight in air Normal to the plane
  14. Feb 23, 2012 #13
    I am just trying to answer the original question. The question you're asking is separate. Regarding projectile motion, which you can read about here:

  15. Feb 23, 2012 #14
    Is the angle given ?
  16. Feb 23, 2012 #15
  17. Feb 23, 2012 #16
  18. Feb 23, 2012 #17
  19. Feb 23, 2012 #18


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    Homework Helper

    Chemicist, your statement is not right.

    The ball is kicked at an angle with the horizontal:

    Initially the ball has both horizontal and vertical velocity components. When in air, the only force acting on the ball is vertical. So the ball keeps its initial horizontal velocity component and the vertical component will change according to v(vertical) =V0(vertical)-gt. At the top of the arc, the vertical velocity component is zero, not the KE.

  20. Feb 23, 2012 #19
    Apply conservation of energy at top taking initial speed v (which is given ) and at the maxima , velocity will be vcos¢ , so you have equation in one variable from where you can find the angle
  21. Feb 23, 2012 #20


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    Homework Helper

    Kushan is right, both the initial speed and height of the ball are given, so you can get the total energy, E, which is conserved. At the top of the arc, the potential energy corresponds to the height of 9.5 m. Subtract it from the total energy, you get the KE there.
    Reaching the ground, all energy is kinetic.

  22. Feb 23, 2012 #21
    Ok I tried redoing the whole problem...

    KE= 36.125 J
    PE = 0 J

    I found the total E to be 36.125 J

    Then i found the PE2 to be 23.31 J (0.250) (9.81) (9.50) = 23.31
    Then subtracted 23.31 from the total E 36.125 - 23.31 = 12.815
    So the KE2 = 12.815 J

    Then I found the speed by calculating (17.0)^2-2(9.81)(9.5)
    V= 10.13 m/s

    KE1 + PE1 = KE2 + PE2
    36.125 = 23.31 + 12.815

    Does this seem correct?

    I also found the speed of the ball v when it reaches the ground to be:
    v^2= 2gh
    v^2= 2 (9.81) (9.50)
    v = 13.65 m/s
    Last edited: Feb 23, 2012
  23. Feb 23, 2012 #22


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    The ball is kicked from 1.5 m height and reaches 9.5 m above the ground. The initial potential energy is not zero. Redo the calculations.

  24. Feb 23, 2012 #23
    chemicist is even wrong that at arc velocity will be zero
    its Horizontal component of Velocity that is NOT ZERO
  25. Feb 23, 2012 #24
    really? if in a space no force is acting on a obbject and it standing still does it have a potential to do work?
  26. Feb 23, 2012 #25
    yea zero potential
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