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Conservation of Energy + Projectile Motion

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data
    So I have a ball rolling down a ramp structure from rest and up a curve,to launch out at 15 degrees. The ball lands 100cm away in a hole 0.05cm deep.
    The height in which it comes out of the ramp is 0.05cm high
    angle at which ball launches out = 15 degrees above horiontal
    dx=100cm
    dy=-0.05+-0.05= -0.1cm
    *also the lowest point of the ramp is on the ground

    find the height in which the ball needs to be rolled down the ramp in order to land 100cm away and into a hole 0.05cm deep.


    2. Relevant equations
    Conservation of energy
    projectile motion equations


    3. The attempt at a solution
    Use conservation of energy equation to find velocity and than input that into dy=vyit+1/2ayt^2
    Ei=Ef
    Ei=KEt+KEr
    mgh=1/2mv^2+1/2Iw^2
    I=2/5mr^2
    w=v/r
    mgh=7/10mv^2
    solve for v, v=sqrt(10/7gh)---= launching velocity
    Use formula dy=vyit+1/2ayt^2
    t=dx/vx
    vx=cos(15)vi
    vyi=sin(15)vi
    -0.1=sin15vi(100/cos15vi)+1/2(-9.81)(1/(cos15)(vi))^2
    vi cancels vi=launching velocity = sqrt(10/7gh)
    -0.1=sin15(100/cos15)+1/2(-9.81)(100^2/(cos15)^2(10/7(9.81)h)
    9.81 cancels
    -0.1=sin15(100/cos15)+1/2(-1)(100^2/(cos15)^2(10/7h)
    solve for h
    h=-70.9cm
    Umm, I don't get why my height becomes negative..so I think I did something wrong but
    I don't know where. ---This is actually from my lab and not a problem so I might be missing something.
     
    Last edited: Jan 6, 2013
  2. jcsd
  3. Jan 6, 2013 #2
    I'm slightly confused about what the 100/cos15 is. Is that your time?
     
  4. Jan 6, 2013 #3
    Um oh sorry about the confusion...I accidently inputed some wrong numbers, give me a second
    and yes that is my time
     
    Last edited: Jan 6, 2013
  5. Jan 6, 2013 #4
    Where does "100" come from?
     
  6. Jan 6, 2013 #5
    So it lands in a hole 100cm* away in 0.05cm* deep hole.
     
    Last edited: Jan 6, 2013
  7. Jan 6, 2013 #6
    You can't mix meters and centimeters in one formula. "9.81" is in meters.
     
  8. Jan 6, 2013 #7
    lol, okay I think I know why my calculations are wrong now...my bad
    TYVM
    One more thing, so am I approaching the calculation/problem correctly? by using conservation of energy to find launch velocity and than input that into the dy=vyit+1/2ayt^2?
     
  9. Jan 6, 2013 #8
    The overall approach seems OK. There might be errors in algebra - the formulas are very hard to read.
     
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