1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  2. jcsd
  3. Aug 19, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.

  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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Resolving max height with calculus
  1. Calculus - Max/Min (Replies: 8)