During liftoff, a hot-air balloon accelerates upward at a rate of 3.4m/s^2. The

Question

During liftoff, a hot-air balloon accelerates upward at a rate of 3.4m/s^2. The balloonist drops an object over the side of the gondola when the speed is 17m/s.
The magnitude of the objects acceleration after it is released is 9.8 m/s^2. And the direction of the acceleration is downward.
How long does it take to hit the ground? (Express your answer using two significant figures) t= __ s During liftoff, a hot-air balloon accelerates upward at a rate of 3.4m/s^2. The balloonist drops an object over the side of the gondola when the speed is 17m/s.
The magnitude of the objects acceleration after it is released is 9.8 m/s^2. And the direction of the acceleration is downward.
How long does it take to hit the ground? (Express your answer using two significant figures) t= __ s
The magnitude of the objects acceleration after it is released is 9.8 m/s^2. And the direction of the acceleration is downward.
How long does it take to hit the ground? (Express your answer using two significant figures) t= __ s

Explanation / Answer

first

t = v / a

t = 17 / 3.4 = 5 sec

the height it has raisen

d = 0.5 * a * t^2

d = 0.5 * 3.4 * 5^2 = 42.5 m

Now you can calculate the additional time t' it takes for the dropped object to hit the ground

by using the second equation of motion

42.5 + 17 * t - 4.9 * t^2 = 0

t = 5.15 sec , -1.68 sec

take the positive value t = 5.15 sec

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