SOLUTION: A rocket is launched from Venus with a velocity of 100 ft/sec. The gravity on Venus is 28.8 ft/sec squared. After 3 seconds, how high will the rocket be? (Use the formula y=1/2g

Question 189083: A rocket is launched from Venus with a velocity of 100 ft/sec. The gravity on Venus is 28.8 ft/sec squared. After 3 seconds, how high will the rocket be? (Use the formula y=1/2gt squared + vt + s where g is the gravity, v is velocity and s is initial height). How long does it take the rocket to land? To reach its maximum height? What is the maximum height of the rocket?

2. A punter in the NFL wants his kick to remain in the air for the longest possible time in order that his defense gets as far down the field as possible. This is called "hang time". If the equation f(h)=-4.9t squared + 22t + 2 models the path of the punt, how long does the fottball remain airborne?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A rocket is launched from Venus with a velocity of 100 ft/sec. The gravity on Venus is 28.8 ft/sec squared. After 3 seconds, how high will the rocket be? (Use the formula y=1/2gt squared + vt + s where g is the gravity, v is velocity and s is initial height). How long does it take the rocket to land? To reach its maximum height? What is the maximum height of the rocket?
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y=1/2gt^2 + vt + s
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Your Problem:
h(t) = (1/2)(-28.8)t^2 + 100t + 0
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h(3) = (-14.4)(3^2) + 100*3
h(3) = 170.4 ft.
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When the rocket lands its height will be zero.
(1/2)(-28.8)t^2 + 100t = 0
-14.4t^2 + 100t = 0
-2t(7.2t - 100) = 0
Positive solution:
7.2t -100 = 0
t = 100/7.2 = 13.89 seconds
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2. A punter in the NFL wants his kick to remain in the air for the longest possible time in order that his defense gets as far down the field as possible. This is called "hang time". If the equation f(h)=-4.9t squared + 22t + 2 models the path of the punt, how long does the fottball remain airborne?
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-4.9t^2 + 22t + 2 = 0
Positive solution:
t = [-22 - sqrt(22^2 -4*-4.9*2)]/(-9.8)
t = 9.8 seconds
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Cheers,
Stan H.

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