A quarterback takes the ball from the line of scrimmage, runs backward for 12.1 yds, then runs sideways parallel to the line of

Question

2 years ago

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A quarterback takes the ball from the line of scrimmage, runs backward for 12.1 yds, then runs sideways parallel to the line of

scrimmage for 19.8 yds. At this point, he throws a 41.5 yd forward pass straight downfield, perpendicular to the line of scrimmage. What is the magnitude of the football's resultant displacement?

Physics

1 answer:

Over [174] · Answer 1 · 2022-07-15T09:07:53+03:00

Answer:

The answer is 35.45 yds

Explanation:

You have to picture this to be able to understand it better (see attachment).

Start at the origin which is when the quarterback (QB) takes the ball. He runs backwards 12.1 yards, runs sideways for 19.8 yards (it doesn´t matter if he runs right or left), then he throws the ball forward 41.5 yards. If you look at the attachment, you can see I drew the path that the football followed. And then connected the dots from the origin and finish. The distance between those two points is the magnitude of the resultant displacement.

In order to calculate it, all you need to do is use the Pythagoream theorem, which says that the square of the hypotenuse equals the sum of the squares of the legs a and b of the triangle rectangle.

$R^{2} = a^{2} + b^{2}$ then solve for R
$R = \sqrt{a^{2}+b^{2} }$

In this case, you know the length of leg a to be 19.8 yards which how much it moves sideways. And then, to get the length of leg b, all you need to do is substract how much it moved backwards from the 41.5yards forward displacement. This results in b leg being 29.4 yards long.

Now you have a triangle with:

a = 19.8 yards
b = 29.4 yards

Substituting this numbers in the equation:

$R = \sqrt{19.8^{2}+29.4^{2} }$
R = 35.45 yards