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Conservation of Energy of a thrown stone

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown upward at an angle of 53° above the horizontal. Its
    maximum height above the release point is 24 m. What was the stone’s initial
    speed? Assume any effects of air resistance are negligible.

    2. Relevant equations
    Emech=0


    3. The attempt at a solution
    On my book's solution, I do not understand, why the V(y) component of the velocity is not taken into account. Also, the solution is not given in polar form, but the V(x) component was used at Hmax.

    My solution used the Y component of the velocity at Hmax and is 85m/s. I used conservation of Emech at initial position and Hmax.
    Thank you.
     
  2. jcsd
  3. Aug 12, 2011 #2

    gneill

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    Staff: Mentor

    Can you elaborate on your solution? We can't comment on what we can't see :smile:
     
  4. Aug 12, 2011 #3
    Ok, V(y) is 0 m/s at max height.

    So, in the Y direction, you have the following

    U (initial velocity) = X
    V (Final velocity) = 0
    A (acceleration) = -9.8 m/s/s
    S (displacement) = 24
    T (time) = t

    Using the appropriate equation of motion ( V^2 = U^2 + 2as) you can then find the Y comonpen of initial velocity.


    BUT YOU'RE ASKED FOR SPEED.

    Remember Speed is a scaler quantity and is typical the magnitude of the vector.

    Please say if you need help finding total time etc..
     
  5. Aug 12, 2011 #4
    DarthFrodo: I need to use Conservation of energy equations.

    Gneill: I thought that I made a straight forward and known mistake.
    My procedure is:

    Position 1 is the ground level. Position 2 is at Hmax.
    Using position 2 and 1 for conservation on energy:
    K2-K1+ U2-U1=0, where K and U are kinetic and gravitational potential energy respectively. Also, there is no air drag or other forces interfering.

    K2 is zero, for I used the V(y) component. This is one aspect of the solution that differs from the book's solution.
    So,

    -(1/2)(Vsin(angle))^(2) +mg(Hmax)=0 , V is initial velocity.

    48g/ (sin(angle))^(2) =(V)^(2)

    (1/sin53)( (48)(9.81) )^(1/2)=V

    85.1022= V.
     
  6. Aug 12, 2011 #5

    SammyS

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    K2 ≠ 0 . The horizontal component of velocity is NOT zero. (V1)x = (V2)x = Vinitial cos(θ)

    (1/2)mV2 = (1/2)m(Vx2 + Vy2)
     
  7. Aug 12, 2011 #6

    gneill

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    Staff: Mentor

    You have a problem with your final value. The second to last line looks okay, but the numerical result that you calculate from it is not! Perhaps you can detail the intermediate values in the calculation?
     
  8. Aug 12, 2011 #7
    On my first post:
    3. The attempt at a solution
    On my book's solution, I do not understand, why the V(y), vertical component of the velocity, is not taken into account.

    Now, I understand the procedure. I did not realize that both components, where taken into account, but not all were written.

    Thank you, SammyS.
     
  9. Aug 12, 2011 #8

    gneill

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    Staff: Mentor

    You're working with the vertical component of the velocity when you write V*sin(53°), where V is the initial speed. You're applying conservation of energy to find this initial vertical component, V*sin(53°), which is fine.
     
  10. Aug 12, 2011 #9
    A side question here:

    What does the knowNothing23 mean by "Emech=0"?
     
  11. Aug 13, 2011 #10

    PeterO

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    Perhaps you have written it in an unusual way, but the vertical component [y-component] is zero at maximum height; otherwise it will be still getting higher?

    The angle - 53° - gives you the relationship between vertical and horizontal components at launch. [That angle is very familiar to regular problem solvers].
    The maximum height reached enables you to calculate the vertical component. Once you have the vertical and horizontal components of V you can get the actual V
     
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