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

by xcgirl
Tags: 60yd, angle, degree, quarterback, throw
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xcgirl
#1
Feb23-11, 01:30 PM
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?
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gneill
#2
Feb23-11, 02:07 PM
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P: 11,617
When the throw has been measured as having gone 60 yards, what does that mean? Is the ball still flying through the air?
xcgirl
#3
Feb23-11, 02:11 PM
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.

gneill
#4
Feb23-11, 02:19 PM
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P: 11,617
A quarterback can throw 60yd @ 45 degree angle. How fast can he throw?

Quote Quote by xcgirl View Post
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...
xcgirl
#5
Feb23-11, 03:16 PM
P: 20
thanks, i got it now!


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