A hot-air balloon is rising upward with a constant speed of 2.13 m/s. When the balloon is 3.14 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
At the moment that the compass is released, it is moving upward at a speed of 2.13 m/s. The velocity will decrease
9.8 m/s each second, until it reaches its maximum height. Then, the compass will fall toward the ground. During the
fall, the velocity increases 9.8 m/s each second.
Final velocity^2 - Initial velocity^2 = 2 * acceleration *
displacement
Let upward motion be positive
vi = +2.13 m/s
a = -9.8 m/s^2
d = -3.14 m
vf^2 - 2.13^2 = 2 * -9.8 * -3.14
vf^2 = 2.13^2 + (2 * -9.8 * -3.14)
vf = ±8.13 m/s
Since the compass is falling down, the velocity is -8.13
m/s
Time = (vf - vi) ÷ a = (-8.13 - 2.13) ÷ -9.8 = 1.05 seconds
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