How Fast Must a Quarterback Throw to Achieve a 171 Meter Pass?

AI Thread Summary
To determine the speed required for a quarterback to throw a football 171 meters at a 26.2° angle, the projectile motion formula R = v²*sin(2θ)/g is used. The horizontal distance (x) is given as 171 meters, with no horizontal acceleration, and the vertical distance (y) is zero. The challenge lies in the need for additional variables to solve for the initial velocity (v). The discussion highlights the importance of understanding the derivation of the range formula in projectile motion. Ultimately, the conversation seeks clarification on the formula's origin and application.
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Homework Statement



A quarterback claims that he can throw the football a horizontal distance of 171 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 26.2 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Homework Equations



vx = v0x + axt
vy = v0y + ayt
x = 1/2(v0x + vx)t
y = 1/2(v0y + vy)t
x=v0xt + 1/2axt2
y=v0yt + 1/2ayt2
vx2 = v0x2 + 2axx
vy2 = v0y2 + 2ayy



The Attempt at a Solution



The difficulty that I'm having is it seems like I don't have enough information to solve the problem. I've got the horizontal distance (x) as 171 meters, the horizontal acceleration (ax) as 0 m/s2, the vertical distance (y) as 0 meters and the vertical acceleration (ay) as -9.80 m/s2.

I know all I need is one more variable in order to solve this problem, but I just can't seem to figure out what I'm missing given the information I've got.
 
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Using the relevant equations, you can show that the range of the projectile is given by
R = v^2*sin(2θ) /g. Using this formula find v.
 
rl.bhat said:
Using the relevant equations, you can show that the range of the projectile is given by
R = v^2*sin(2θ) /g. Using this formula find v.

Would you mind explaining how you came up with that formula?
 
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