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Resolving max height with calculus

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown at 32m/s vertically what is the max height the ball can attain.


    2. Relevant equations
    no equation is given except initial speed


    3. The attempt at a solution
    so my guess is since the gravity is 9.8m/s^2

    if I integrate 9.8m/s^2 = 32 it gives me 2.1396
    so after 2.1396 the ball should have its max height.


    So I end up with this equation
    height = 32*2.1396 - 9.8*2.1396^2
    height = 23.60 meters

    But I am soo uncertain about my answer, I have to use calculus to resolve it

    Thank you for your help

    Jonathan
     
  2. jcsd
  3. Aug 19, 2012 #2

    LCKurtz

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    Start out with the equation for acceleration: ##a=-9.8##. Integrate that with respect to time twice. What do you get for velocity ##v## and position ##s##? Don't forget to think about the constants of integration.
     
    Last edited: Aug 19, 2012
  4. Aug 19, 2012 #3
    What can't you use the equation for kinetic/potential energy?
     
  5. Aug 19, 2012 #4
    no I can't use the kinect equation, but i found the answer with calculus

    f(0) = 32 so
    speed = f(x) = -9.8x + 32
    so if I integrate this from 0 to 32/9.8 it gives me 52.24 meters

    (-9.8t^2)/2 + 32x
    (-9.8(32/9.8)^2)/2 + 32(32/9.8) = 52.24 meters

    which is a mere aproximate
     
  6. Aug 19, 2012 #5

    LCKurtz

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    Your answer is correct but you could certainly improve the writeup. I will insert suggested improvements in red.

    ##s(t)=\frac {-9.8t^2}{2}+32t##
    ##s(\frac{32}{9.8})=52.24##


    Even with those improvements, you have hand-waved evaluating the constants and evaluating the ##t## value for max height, which you seem to understand.
     
  7. Aug 19, 2012 #6
    Thank you for your help

    I had no idea how to type correctly on the forum
     
  8. Aug 19, 2012 #7

    LCKurtz

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    Of course, proper notation is more than typing. For example mixing up ##x## and ##t## in your equations.

    Also, voko's suggestion for using KE/PE equations gives an easy solution. At the bottom the kinetic energy is ##\frac 1 2 mv^2## and at the top there is no KE but the potential energy is ##mgh##. Setting those equal and solving for ##h## gives the same answer.
     
  9. Aug 19, 2012 #8

    Ray Vickson

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    He was not referring to the use of LaTeX, etc., but, rather, to the method of presentation. There would be nothing wrong in writing s(t) = -(1/2)(9.8)t^2 - 32t, or (1/2)*(9.8)*t^2 - 32*t for example, instead of [tex] s(t) = -\frac{1}{2} 9.8 t^2 - 32 t.[/tex] As long as you use brackets to make things clear, you should be OK---but do not "overuse" them either.

    Generally, in Physics especially, it is a bad idea to use "x" or "y" to refer to time (even though time may be an unknown in your problem); usually we employ the letters "x" or "y" to refer to coordinates on a line, or in a plane or in space. Of course, using "x" as a time variable is not illegal, but it is unwise; certainly, you should use the same notation from start to finish of a problem (which you did not: you wrote (-9.8t^2)/2 + 32x, where x and t are supposed to be the same thing!) Writing up your work to make it easier to mark is always recommended.

    RGV
     
  10. Aug 20, 2012 #9
    Perhaps because he's never heard of it. The fact that he's posting in the calculus section suggests that the problem is from a math book, not a physics book. And even in most freshman physics books, KE/PE concepts are introduced several chapters later than the kind of problem in the OP.
     
    Last edited: Aug 20, 2012
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