A hot-air balloon is rising straight up with a speed of 3.15 m/s. A ballast bag is released from rest relative to the balloon when it is 6.89 m above the ground. How much time elapses before the ballast bag hits the ground?
0=3.15 - gt
t= 3.15/9.8 =0.32 sec
0 = (3.15)^2 - 2*9.8*x
x=0.5m
hight = 6.89 + (0.5) = 7.4 m
7.4 = (1/2)*9.8*t^2
t=1.23 s
net time = 1.23 + 0.32 =1.55 sec
A hot-air balloon is rising straight up with a speed of 3.15 m/s. A ballast bag...
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