Question

[[[Physics]]]

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The football team at Enormous State University (ESU) uses vector displacements to record its plays, with the origin taken to be the position of the ball before the play starts. In a certain pass play, the receiver starts at +1.0i^−5.0j^+1.0i^−5.0j^, where the units are yards, i^i^ is to the right, and j^j^ is downfield. Subsequent displacements of the receiver are +9.0i^+9.0i^ (he is in motion before the snap), +11.0j^+11.0j^ (breaks downfield), −6.0i^+4.0j^−6.0i^+4.0j^ (zigs), and +12.0i^+18.0j^+12.0i^+18.0j^ (zags). Meanwhile, the quarterback has dropped straight back to a position −7.0j^−7.0j^.

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(1)

How far must the quarterback throw the ball? (Like the coach, you’ll be well advised to diagram the situation before solving this numerically.) Express your answer in yards.

(2)

In which direction must the quarterback throw the ball?

Express your answer answer in degree. (degree to the right of down field)

Answer 1

Part A:

let the quadrack throw the ball to S yards

To develop this problem we will apply the concepts related to the vector sum. For this purpose, we will take the displacement vectors of the individual and add them to obtain the total distance traveled by him, from the 0 position, to the final position.

so let the displacement vector is

=(1.0i^−5.0j^)+(9.0i^+11.0j^)+(−6.0i^+4.0j^)+( +12.0i^+18.0j^)

=16i^+28j^

With the quarterback's position

=-7j^

Now we can find the relative position of the two individuals through the subtraction of the vectors, therefore the position between the individual and the quarterback would be

=-

=(16i^+28j^)-(-7j^)

=16i^+35j^

Through the properties of the vectors the value of the magnitude of the relative distance would be

=38.48 yards.

partB:

The direction of the vector would be given by the tangent of the two vectors, that is

=(35/16)

=

Therefore the quarterback will throw the ball 44.4 yards in one direction, with respect to