Asked by ali
A quarterback throws a football with a velocity vo at an angle of 45o with the horizontal. At the same instant a receiver standing 20 ft in front starts running down the field at 15 ft/s and catches the ball. What is the distance of the receiver from the quarterback when the ball is caught. Assume the ball is thrown and caught at the same height above the ground.
Answers
Answered by
drwls
distance ball thrown = receiver's distance from QB when caught
[2 vo sin 45/g]*vr + 20 = vo^2/g
vr = runner's velocity = 15 ft/s
g = 32.2 ft/s^2
You have one unknown, vo. Solve for it (using the quadratic formula) and use the answer to compute the distance thrown, vo^2/g.
[2 vo sin 45/g]*vr + 20 = vo^2/g
vr = runner's velocity = 15 ft/s
g = 32.2 ft/s^2
You have one unknown, vo. Solve for it (using the quadratic formula) and use the answer to compute the distance thrown, vo^2/g.
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