Question

By how much does the ball clear or fall short of clearing the crossbar??

Answer 1

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.