SOLUTION: A balloon rising vertically with a velocity of 16 ft/sec, releases a sandbag at an instant when the ballon is 64 ft. above the ground. Compute the position of the sandbag 1/4 sec.

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Question 895481: A balloon rising vertically with a velocity of 16 ft/sec, releases a sandbag at an instant when the ballon is 64 ft. above the ground. Compute the position of the sandbag 1/4 sec. after its release.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The formula is

h=h%5B0%5D%2Bv%5B0%5Dt-16t%5E2

h = final height in feet
h0 = the initial height in feet
v0 = the initial velocity in ft/sec
t = the final time.

h = what we are looking for.
h0 = 64 ft
v0 = 16 ft/sec
t = 1/4 sec

h=h%5B0%5D%2Bv%5B0%5Dt-16t%5E2
h=64%2B16%281%2F4%29-16%281%2F4%29%5E2
h=64%2B4-16%281%2F16%29
h=68-1
h=67

Answer: 67 feet above the ground

[Notice: Since the balloon was going up when the
sandbag was released, the sandbag was still going up
1/4 of a second after it was released.  It had not
started to fall yet.] 

Edwin