1. Not finding help here? Sign up for a free 30min 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!

What speed must the ball be thrown

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    What speed must a ball be thrown vertically to reach a max height of 70m?

    2. Relevant equations



    3. The attempt at a solution

    I originally tried to find the velocity by the fact that the velocity of the ball must be zero at a height of 70 m, accounting for 9.8 m/s^2 acceleration due to gravity (which is negative in this case)

    So, if the velocity is decreased by 9.8 m/s every second, and reaches 70 m at zero, then I could find that if I knew the time that the ball is in the air.

    To find the time, I could use x=xo + vot + (1/2)at^2, but then I need initial velocity!

    What formula am I missing?
     
  2. jcsd
  3. Aug 24, 2011 #2
    Try using the first equation of motion.
    v=v0+at.
    You know that at the maximum height the velocity is zero, using that you can find the value of v0 (in terms of g and t).
    Then substitute it in second equation of motion. ;)
     
  4. Aug 24, 2011 #3
    I'm not sure I understand how I could apply that one. I have v, as 0, but the only other value I know is a at 9.8.

    0 = Vo + 9.8t

    What can that tell me?

    Thanks!
     
  5. Aug 24, 2011 #4
    Oops!! You did a sign mistake :smile:
    What you should get is this:-
    0=v0-gt

    For now, don't substitute g, when the equations are solved then you can substitute g.
     
  6. Aug 24, 2011 #5
    So Vo=gt. So replace all instances of Vo with gt in the time equation, right?

    Thanks!
     
  7. Aug 24, 2011 #6
    Yes! :)
     
  8. Aug 24, 2011 #7
    Do you agree with t = 2.18?
     
  9. Aug 24, 2011 #8
    My answer comes out to be 3.74s when g=10 m/s^2.
    It dont come 2.18s too when i take g=9.8 m/s^2
     
  10. Aug 24, 2011 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Since Vo is what you want to solve for, not really the time, wouldn't it make more sense to eliminate t instead?
     
  11. Aug 24, 2011 #10
    Going that way I get vo= 21.4. Sorry I am showing no work, on a cellphone.
     
  12. Aug 24, 2011 #11

    Doc Al

    User Avatar

    Staff: Mentor

    No, but I have no idea what you did wrong.
     
  13. Aug 24, 2011 #12
    Alright, guess ill just have to wait until I get home, sit down, and think about it.
     
  14. Aug 24, 2011 #13
    Here's my work for replacing Vo

    Given that Xo = 0, X = 70, Vo = gt

    [itex]x = x_{0} + v_{0}t + \frac{1}{2}gt^{2}[/itex]

    Replace instances of Vo with gt, Xo is 0. x is 70.


    [itex]70 = (gt)t + \frac{1}{2}gt^{2}[/itex]


    [itex]70 = gt^{2} + \frac{1}{2}gt^{2}[/itex]


    [itex]70 = 9.8t^{2} + 4.9t^{2}[/itex]


    [itex]70 = 17.4t^{2}[/itex]

    [itex]4.76 = t^{2}[/itex]

    t = 2.18
     
  15. Aug 24, 2011 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Your last term has the wrong sign. The acceleration is -g, not g.
     
  16. Aug 24, 2011 #15
    Of course! Ball is thrown against gravity. Accounting for that, 3.77964473..

    Thanks a lot guys.
     
  17. Aug 24, 2011 #16
    37.42 for initial velocity?
     
  18. Aug 25, 2011 #17
    Yep you got it. :smile:
     
  19. Aug 25, 2011 #18

    Doc Al

    User Avatar

    Staff: Mentor

    Good. Now try doing as I suggest in post #9 and solve it without the unnecessary step of calculating the time.
     
  20. Aug 25, 2011 #19
    It wasn't unnessesary because part b asked for it anyway, but will do.
     
  21. Aug 25, 2011 #20

    Doc Al

    User Avatar

    Staff: Mentor

    It's worth doing. And in the process you'll derive a very useful kinematic formula that will enable you to solve for the velocity in one step. (Well worth committing to memory.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook