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

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A quarterback can throw a football 60 yards at a 45-degree angle, which translates to 54.86 meters. To solve for the throw speed, the equations of motion in both x and y directions are used, with the x-direction equation being d = vt and the y-direction equation accounting for gravity. The net displacement in the y-direction is zero since the ball is caught at the same height from which it was thrown. By substituting this into the y-direction equation, the time variable can be expressed in terms of other variables. This approach allows for the calculation of the initial throw speed needed to achieve the distance.
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Homework Statement



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

Homework Equations



60 yds = 54.86 meters

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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When the throw has been measured as having gone 60 yards, what does that mean? Is the ball still flying through the air?
 
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.
 
xcgirl said:
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...
 
thanks, i got it now!
 
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