A quarterback can run with a speed of vq = 15 mph. He throws a ball with a speed of vb = 35 mph at some unknown angle θ between 30 and 60 degrees, which is measured from horizontal. Neglect air resistance.- a reciever runs past the quarteback ( just beside him) at speed vq. If the quarterback releases his pass at that instant what would be the angle in degrees that he would have to throw the ball for the reciever to catch it?

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Asked Sep 14, 2019
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A quarterback can run with a speed of vq = 15 mph. He throws a ball with a speed of vb = 35 mph at some unknown angle θ between 30 and 60 degrees, which is measured from horizontal. Neglect air resistance.

- a reciever runs past the quarteback ( just beside him) at speed vq. If the quarterback releases his pass at that instant what would be the angle in degrees that he would have to throw the ball for the reciever to catch it?

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Expert Answer

Step 1

the reciever can catch the ball if and only if the horizontal component of speed of the ball is equal to the speed of the reciever.

Step 2

equate the the horizontal componet speed of the ball and speed of the reciever.

= cose
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= cose

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Step 3

substitute the given quantities in the abo...

cose
e cos
15mph
cos
35mph
=64.6
65°
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cose e cos 15mph cos 35mph =64.6 65°

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