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!

Homework Help: Projectile problem w/ unkown velocity/time

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    An athlete executing a long jump leaves the ground at a 25° angle and lands 6.30 m away.
    What was his take off speed?
    2. Relevant equations
    (In all equations, X can be substituted for Y in entire equation)
    Vy= Voy + AyT
    Vy^2 = Voy + 2AyY
    Y=VoyT + (1/2)AyT^2

    3. The attempt at a solution
    The answer is 8.98 m/s. I can't figure out how to do this problem, and need to understand this for an upcoming final. I have worked several other similar problems, but none without a known velocity and/or time.
    Any insight into how to even begin would be greatly appreciated.
    Last edited: Dec 8, 2008
  2. jcsd
  3. Dec 8, 2008 #2
    Sorry I edited this because I think i made a mistake don't worry I didn't give up on you.

    I'm sorry again I keep getting an answer very close and is probably just off because of my rounding. Though it's been a while since I have done projectile motion and I would hate to show you the wrong way with a final coming up.
    Last edited: Dec 8, 2008
  4. Dec 8, 2008 #3


    User Avatar
    Homework Helper

    The x component of velocity is Vo*Cosθ

    His total distance would be given by Vo*Cosθ*t

    So ...

    X = 6.3m = Vo*Cos25*t

    Now you still need time so ... look to the skies.

    Vo*Sinθ = g*t and t to max height = Vo*Sinθ/g

    But total time = 2*t

    Now we can solve. t = 2*Vo*Sin25/g = 6.3/Vo*Cos25

    Rearranging Vo2*2*Sin25*Cos25 = 6.3*g

    Recognizing that 2*Sinθ*Cosθ = Sin2θ

    Vo2 = 6.3*g/Sin50
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook