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Free Fall stone physics

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown upward at an angle of 51° above the horizontal. Its maximum height during the trajectory is 26 m. What was the stone's initial speed?
    m/s

    2. Relevant equations

    a=v2/r
    v2=vo2+2ax
    C=2[tex]\pi[/tex]r

    3. The attempt at a solution

    I figured the acceleration is 9.81*sin51 since it just gravity when thrown upward so now I have to account for gravity.
    so a=7.62

    I plugged this into a=v2/r in order to find velocity:
    7.62=v2/26
    so v=14.079

    The distance I got by using the equation C=2[tex]\pi[/tex]r
    I got that C=163.36 and the distance is 1/4th of the circumference since the ground and highest point are perpendicular. I got 40.84 for the distance

    I then plugged my answers for a, v and distance into v2=vo2+2ax to find initial velocity
    I got that v=20.596 but this was incorrect. Does anyone know what I did wrong?
    Thank you!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 22, 2008 #2

    mgb_phys

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    Almost correct approach, consider the vertical component of the motion.
    You know the vertical accelration and maximum height - you can then work out the vertical component of the initial speed.
    Alternatively you can use conservation of energy.
     
  4. Oct 22, 2008 #3
    Okay so the vertical component of acceleration would be 9.81*cos51 right?

    And then plug this into v2=vo2+2ax using 26 for x?

    And how do I get the answer from the vertical and horizontal components of the initial speed?
     
  5. Oct 22, 2008 #4
    No! Draw a picture and think first!
     
  6. Oct 22, 2008 #5
    Moreover, your equations C=2\pi r and a=v^2/r are irrelevant: the stone does not follow a circular path
     
  7. Oct 22, 2008 #6
    Okay thank you. I tried a new approach using the conservation of energy and the formula Ki+Ui=Kf+Uf.
    This would be 1/2mVi^2+mgh=1/2mVf^2+mgh.
    1/2mVi^2+m(9.81 cos 51)*0=1/2*0^2+m(9.81 cos 51)*26
    1/2mVi^2=m(9.81 cos 51)*26
    So after the masses cancel out, Vi should be 17.9 m/s
    Did I do this correctly?
     
  8. Oct 22, 2008 #7

    mgb_phys

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    You have to lose this idea the the acceleration is cos(51).
    Vertically the accelration is 9.81m/s^2 downward (or -9.81m/s^2 up if you prefer)
    The horizontal accelaration is zero.
     
  9. Oct 22, 2008 #8
    So I don't need to account for the angle any where in the problem?

    Other than the acceleration is there anything else wrong my work?
     
  10. Oct 22, 2008 #9

    mgb_phys

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    You need to account for the angle in converting the vertical and horizontal velocities back into a velocity along the line for the final answer.

    A really nice feature about forces is that you can split them up into components, deal with them separately and simply add them back together correctly.
    So in this case we simplify it by looking at the vertical component where you know the final velocity, accelration and distance. If you had been given the total horizontla distance and time you could work out the speed using the horizontal part and ignore the vertical motion completely.

    Most people find conservation of energy simpler to understand than forces - and it's much harder to get the signs wrong.
     
  11. Oct 22, 2008 #10
    Okay thank you. so if I use the formula Ki+Ui=Kf+Uf I don't need to multiply gravity by cos(angle)?
     
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