SOLUTION: A launcher is used to shoot water balloons up into the air. The initial velocity of the water balloon affects both how high it goes and how long it stays in the air. How many seco
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Question 821976: A launcher is used to shoot water balloons up into the air. The initial velocity of the water balloon affects both how high it goes and how long it stays in the air. How many seconds will people have to dodge the balloon if it is launched with
a. Initial velocity 320 ft./sec?
b. Initial velocity 240 ft./sec?
c. Initial velocity 160 ft./sec?
d. Initial velocity 80 ft./sec?
e. Initial velocity 64 ft./sec?
f. Initial velocity 32 ft./sec?
Thanks in advance
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A launcher is used to shoot water balloons up into the air. The initial velocity of the water balloon affects both how high it goes and how long it stays in the air. How many seconds will people have to dodge the balloon if it is launched with
a. Initial velocity 320 ft./sec?
b. Initial velocity 240 ft./sec?
c. Initial velocity 160 ft./sec?
d. Initial velocity 80 ft./sec?
e. Initial velocity 64 ft./sec?
f. Initial velocity 32 ft./sec?
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You didn't spec a function for the height vs. time.
On Earth, h(t) = -16t^2 + v*t + h0 is commonly used, where v = launch speed & h0 = launch height, zero for all these, since no height was given.
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a. Initial velocity 320 ft./sec?
h(t) = -16t^2 + v*t + h0
Solve for t when h(t) = 0
h(t) = -16t^2 + v*t + h0 = 0
h(t) = -16t^2 + 320*t = 0
-16t^2 + 320*t = 0
t = 0 at launch
-16t + 320 = 0
t = 20 seconds at impact.
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Do the others the same way.
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b. Initial velocity 240 ft./sec?
c. Initial velocity 160 ft./sec?
d. Initial velocity 80 ft./sec?
e. Initial velocity 64 ft./sec?
f. Initial velocity 32 ft./sec?
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