SOLUTION: Throwing with the wind, a quarterback passes a football 90 ft in 2 s. Against the wind, the same pass would have traveled 70 ft in 2 s. Find the rate of the pass and the rate of th
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Question 1195078: Throwing with the wind, a quarterback passes a football 90 ft in 2 s. Against the wind, the same pass would have traveled 70 ft in 2 s. Find the rate of the pass and the rate of the wind.
Found 3 solutions by Alan3354, Theo, ikleyn:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Throwing with the wind, a quarterback passes a football 90 ft in 2 s. Against the wind, the same pass would have traveled 70 ft in 2 s. Find the rate of the pass and the rate of the wind.
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Wind affects projectiles, but it cannot be added and subtracted as this problem implies.
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If you toss the ball straight up, it does not move at the airspeed.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
rate * time = distance.
let p = the rate of the pass.
let w = the rate of the wind.
with the wind, the formula becomes:
(p + w) * 2 = 90
against the wind, the formula becomes:
(p - w) * 2 = 70
simplify both equations to get:
2p + 2w = 90
2p - 2w = 70
add the second equation to the first to get:
4p = 160
solve for p to get:
p = 40
the rate of the pass is 40 feet per second.
with the wind, you get:
(40 + w) * 2 = 90
simplify to get:
80 + 2w = 90
solve for w to get:
w = 5
against the wind, you get:\
(40 - w) * 2 = 70
solve for w to get:
w = 5
you get:
the rate of the pass is 40 feet per second.
the rate of the wind is 5 feet per second.
that's your solution.
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
From the Physics point of view, the problem is posed incorrectly and, THEREFORE, makes no sense.
A football is not carried away by the airflow.
Alan said the same by other words.
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Being with this forum several years, I noticed that incorrectly posed problems, similar to this one (of precisely this type)
come to the forum practically each year, with the frequency/periodicity approximately one per year.
So, we, the tutors, give explanation/explanations, that the problem is posed incorrectly,
and it has a validity period about one year, and then it repeats again, with the regularity of wall clocks.
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