A football punt is launched at an angle of 62.5 degrees and has a range of 52.0

Question

A football punt is launched at an angle of 62.5 degrees and has a range of 52.0 m when there is no wind. Suppose the same punt is kicked into a headwind that results in a constant horizontal acceleration of 0.50 m/s2 in a direction opposite to the initial horizontal velocity of the ball.

a)  At the highest point of the trajectory, what is its velocity?

b)  At the highest point of the trajectory, what is its acceleration?

c)  What is the new range of the punt?

d)  Sketch the trajectories of the punt with and without the headwind.

Explanation / Answer

a)

D = v0^/g sin(2 theta)

v0^2 = gD/sin(2theta) = 9.8*52/(sin(2*62.5))

v0 = 24.94 m/s

t = v0 sin(62.5)/g = 24.94/9.8*sin(62.5) = 2.257 s

vx = v0 cos(62.5) - a t = 24.94*cos62.5o - 0.5*2.257 = 10.4 m/s

b)

ax = 0.5

ay = 9.8

a = sqrt(0.5*0.5+9.8*9.8) = 9.81 m/s2

c)

R = 0.5 a (2t)^2 + v0 cos(62.5) (2t) = 0.5*(-0.5)*(2*2.257)*(2*2.257) + 24.94*62.5o * (2*2.257) = 48.9 m

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