From her bedroom window a girl drops a water-filled balloon to the ground, 3.3 m below. If the balloon is released from rest, how long is it in the air?
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To find the time the water-filled balloon is in the air, we can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
h = height (3.3 m)
u = initial velocity (0 m/s since the balloon is released from rest)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time
Since the balloon is dropped from rest, the initial velocity (u) is 0. Therefore, the equation becomes:
h = (1/2)gt^2
Rearranging the equation, we have:
2h = gt^2
Substituting the values:
2(3.3) = (-9.8)t^2
Simplifying:
6.6 = -9.8t^2
Dividing by -9.8:
t^2 = 6.6 / (-9.8)
t^2 ≈ -0.673
Since time cannot be negative in this context, we can conclude that there is no real solution. Therefore, the calculation suggests that the water-filled balloon does not reach the ground, and this can be explained by considering air resistance and other factors not taken into account in our simplified calculation.
In practice, the balloon will experience air resistance, which will slow it down and result in a longer time for it to reach the ground.