In a game of American football, a placekicker must kick a football from a point

Question

In a game of American football, a placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal (distance measured along the ground), and half the crowd hopes the ball will clear (that is, go above) the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 46.0° to the horizontal.

(a) By how much does the ball clear or fall short of clearing the crossbar? (That is, calculate the height of the ball when it gets to the crossbar, and compare to the height of the crossbar itself. Enter a negative answer if it falls short.)

Explanation / Answer

The problem can be solved as follows:

Given d=36.0m, y=3.05m, =46.0, vo=20.0 m/s
Assuming no air resistance, the equations are:
x(t) = vcos()t
y(t) = vsin()t-(g/2)t²

Find time when y(t)=3.05 by using quadratic equation:
(g/2)t²-vsin()t+y(t) = 0
a=g/2, b=-vsin(), c=y
t = (-b±(b²-4ac))/2a
t = (vsin+(v²sin²()-2gy))/g

Plugging this into x(t) above gives:
x(t) = vcos(vsin+(v²sin²()-2gy))/g = 13.89(14.39 + 12.13)9.81 = 37.55 m

So ball clears crossbar by 37.55-36 = 1.55 m

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