Question 287529
 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.