From her bedroom window a girl drops a water-filled balloon to the ground, 4.7 m below. If the balloon is released from rest, how long is it in the air?

3.11 s

To determine how long the water-filled balloon is in the air, we can use the principles of projectile motion. The key idea is that the time it takes for an object to fall to the ground (neglecting air resistance) is the same as the time it takes for an object to be projected upwards at the same initial velocity.

First, let's find the time it takes for the balloon to fall to the ground. We can use the equation:

h = (1/2) * g * t^2

Where,
h = height (4.7 m)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (unknown)

Rearranging the equation, we have:

t^2 = (2h) / g

Substituting the given values, we get:

t^2 = (2 * 4.7) / 9.8

t^2 = 0.959 sec^2

To find the time, we take the square root of both sides:

t ≈ √(0.959)

t ≈ 0.979 seconds

Therefore, it takes approximately 0.979 seconds for the water-filled balloon to fall to the ground.

Since the time it takes for the balloon to reach its highest point is the same as the time it takes to fall to the ground, the total time in the air will be twice the calculated time:

Total time in the air ≈ 2 * 0.979

Total time in the air ≈ 1.958 seconds

Hence, the water-filled balloon is in the air for approximately 1.958 seconds.