OSU\'s football team needs to score a field goal to beat OU. The ball is kicked

Question

OSU's football team needs to score a field goal to beat OU. The ball is kicked (at ground level) from the 21 yard line. In football, the goal posts are an additional distance of 10 yards from the beginning of the end-zone. The ball is kicked at an angle of 60degree to the horizontal and it passes over the goal post which is a vertical distance 10 feet (3.0 m) above the ground (the height of the goal's horizontal crossbar). Ignore air resistance. (1 yard = 0.9144 m) Calculate the minimum speed at which the football is kicked for a successful field goal. Using the speed you found above, Calculate the time that the ball spends in the air. Make sure you calculate for the whole trajectory, kicking to landing at ground level assuming no net to stop the ball.

Explanation / Answer

here,

as 1 yard = 0.9144m
Total Range = 21 + 10 = 31 yards = (31*0.9144) m
angle at which football is kicked, A = 60 degrees

Part A :
Total Distance travelled by projectile is given as :
R = v^2*Sin(2A)/g

Solving for Velocity, v
v^2 = (R*g) / Sin(2A)
v^2 = ( (31*0.9144)*9.8 ) / Sin(2*60)
v = sqrt(320.770) m/s
v = 17.91 m/s or 18 m/s(Rounded off)

Part b :
Time of trajectory will be given as :
T = (2*v*SinA) / g
T = (2*17.91*Sin60)/9.8
T = 3.165 s

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