Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problems with Velocity and Acceleration

  1. Feb 22, 2006 #1
    A ball thrown straigh upward from the ground reaches a maximum height of 400ft. What was its initial velocity? Using the formulas:
    the distance travelled is 400ft, and the initial distance was 0. The force of gravity is -32ft/s/s. Also, the velocity of the ball at the top of the height is 0.

    in the equation V=at+Vo, I don't know how to solve it with 2 variables (t and Vo)
    Last edited: Feb 22, 2006
  2. jcsd
  3. Feb 22, 2006 #2
    V=at+Vo ---------------(A)
    X=1/2at^2+Vot+Xo -----------(B)

    ok. couple of things to note

    assume we are measuring height from ground. (i.e. X0 = 0)
    at maximum height, the velocity of the ball becomes zero for a moment just before it turns around. so at the point V = 0

    so equation (A) becomes
    0 = at + V0 => t = - V0/a

    here 'a' is gravitational acceleration. u need to put correct sign as well.

    i.e. a = -(g). that way your time is positve. (time cannot be negative)

    now substitute this in to (B) and solve for V0.
  4. Feb 22, 2006 #3
    thanks but I'm still stucki got Vo^2=12800/-15 and you can't do a square root of a negative :S
  5. Feb 22, 2006 #4
    thank you so much, i got it. except i made Vo=32t instead. :D
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook