A quarterback can throw 60yd @ 45 degree angle. How fast can he throw?

  1. 1. The problem statement, all variables and given/known data

    A quarterback can throw 60yd @ 45 degree angle. How fast can he throw?

    2. Relevant equations

    60 yds = 54.86 meters

    3. The attempt at a solution

    x-direction
    d = vt
    4.86 = vcos(45)(t)

    y-direction
    d = vt + 1/2at^2
    d = vsin(45)t + 1/2(9.8)t^2

    I want to solve for t in both equations, set them equal, and use that to solve for v but I don't know what to use for d in the y-direction equation. would it be zero because it ends up on the ground?
     
  2. jcsd
  3. gneill

    Staff: Mentor

    When the throw has been measured as having gone 60 yards, what does that mean? Is the ball still flying through the air?
     
  4. I think that it means that the ball is caught at the same height that it is thrown from, so its a net displacement in the y-direction is 0.
     
  5. gneill

    Staff: Mentor

    Okay, so if you put that net y-displacement into your equation of motion for the y-direction, you should be able to solve for t in terms of the other variables in that equation.

    Oh, and make sure that you account for the fact that the acceleration due to gravity is in the downward direction...
     
  6. thanks, i got it now!
     
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