Question 287529: Use a(t) = -32 ft/sec^2 as the acceleration due to gravity. (Neglect air resistance.)
A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 48 feet above the ground.
a) How many seconds after its release will the bag strike the ground?
b) At what velocity will it strike the ground?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Use a(t) = -32 ft/sec^2 as the acceleration due to gravity. (Neglect air resistance.)
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Integrate to get v(t) = -32t + 48 for the velocity
Integrate to get s(t) = -16t^2 + 48t for the position
A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 48 feet above the ground.
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a) How many seconds after its release will the bag strike the ground?
Height is zero when it strikes the ground:
-16t^2+46t+48 = 0
Divide thru by -2 to get:
8t^2 - 23t - 24 = 0
Use the quadratic form to get
positive solution: t = 3.69 seconds
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b) At what velocity will it strike the ground?
v(3.69) = -32(3.69)+48 = -70 ft/sec
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Cheers,
Stan H.
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