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

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

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Answer 1 · 2018-02-28T17:45:39

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:

$x (t) = (22 m/s) cos (47^{\circ}) t - 36 m [1]$ $x(t)= (22" m/s")cos(47^@)t-36" m"" [1]"$

$y (t) = (- 4.9 {m/s}^{2}) t^{2} + (22 m/s) sin (47^{\circ}) t [2]$ $y(t)= (-4.9" m/s"^2)t^2+ (22" m/s")sin(47^@)t" [2]"$

It is easy to solve equation [1] for t because, we have made the goal post at $x (t) = 0$ $x(t) = 0$ :

$0 = (22 m/s) cos (47^{\circ}) t - 36 m$ $0= (22" m/s")cos(47^@)t-36" m"$

$t = \frac{36 m}{(22 m/s) cos (47^{\circ})}$ $t = (36" m")/((22" m/s")cos(47^@))$

$t \approx 2.399 s$ $t ~~ 2.399" s"$

Substitute this value into equation [2]:

$y (2.399 s) = (- 4.9 {m/s}^{2}) {(2.399 s)}^{2} + (22 m/s) sin (47^{\circ}) (2.399 s)$ $y(2.399" s")= (-4.9" m/s"^2)(2.399" s")^2+ (22" m/s")sin(47^@)(2.399" s")$

$y (2.399 s) \approx 10.39 m$ $y(2.399" s") ~~ 10.39" m"$

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.

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

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