SOLUTION: 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 th

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Question 287344: 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 Alan3354(69443) About Me  (Show Source):
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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.
The height of the bag as a function of time in seconds is:
h(t) = -16t^2 + 16t + 48
--------------------------
a) How many seconds after its release will the bag strike the ground?
Find t when h(t) = 0
-16t^2 + 16t + 48 = 0
-t^2 + t + 3 = 0
(-t - 1)*(t -3) = 0
t = 3 seconds
Ignore the t = -1 second
----------------
b) At what velocity will it strike the ground?
vertical speed after release = 16 - 32t
= 16 - 96
= -80 ft/sec (negative is the down direction)