Choosing the x coordinate of the point of the kick to be -36" m"−36 m so that the goal post is at 0" m"0 m, the functions of x and y with respect to time are:
x(t)= (22" m/s")cos(47^@)t-36" m"" [1]"x(t)=(22 m/s)cos(47∘)t−36 m [1]
y(t)= (-4.9" m/s"^2)t^2+ (22" m/s")sin(47^@)t" [2]"y(t)=(−4.9 m/s2)t2+(22 m/s)sin(47∘)t [2]
It is easy to solve equation [1] for t because, we have made the goal post at x(t) = 0x(t)=0:
0= (22" m/s")cos(47^@)t-36" m"0=(22 m/s)cos(47∘)t−36 m
t = (36" m")/((22" m/s")cos(47^@))t=36 m(22 m/s)cos(47∘)
t ~~ 2.399" s"t≈2.399 s
Substitute this value into equation [2]:
y(2.399" s")= (-4.9" m/s"^2)(2.399" s")^2+ (22" m/s")sin(47^@)(2.399" s")y(2.399 s)=(−4.9 m/s2)(2.399 s)2+(22 m/s)sin(47∘)(2.399 s)
y(2.399" s") ~~ 10.39" m"y(2.399 s)≈10.39 m
Subtract 3.05" m"3.05 m to find the distance above the crossbar:
d = 10.39" m" - 3.05" m"d=10.39 m−3.05 m
d = 7.34" m"d=7.34 m above the crossbar.
