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