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

by xcgirl
Tags: 60yd, angle, degree, quarterback, throw
 P: 20 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?
 Mentor P: 11,677 When the throw has been measured as having gone 60 yards, what does that mean? Is the ball still flying through the air?
 P: 20 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.
Mentor
P: 11,677
A quarterback can throw 60yd @ 45 degree angle. How fast can he throw?

 Quote by xcgirl 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.
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...
 P: 20 thanks, i got it now!

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