From her bedroom window a girl drops a water-filled balloon to the ground, 6.9 m below. If the balloon is released from rest, how long is it in the air?
d = (1/2) g t^2
6.9 = 4.9 t^2
To determine the time the balloon takes to reach the ground, we can use the equation of motion for an object in free fall. The equation is as follows:
h = (1/2) * g * t^2
Where:
h = the height of the object (6.9 m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth)
t = time it takes for the object to fall
Now we can rearrange the equation to solve for time (t):
t = sqrt(2h / g)
Substituting the given values, we have:
t = sqrt(2 * 6.9 / 9.8)
t = sqrt(13.8 / 9.8)
t = sqrt(1.408)
t ≈ 1.19 s
Therefore, the balloon will be in the air for approximately 1.19 seconds before reaching the ground.
To find the time for the water-filled balloon to reach the ground, we can use the concept of free fall and the equations of motion. We know that the initial velocity of the balloon is 0 m/s, as it is released from rest. The acceleration due to gravity, represented by "g," is approximately 9.8 m/s² (varies depending on location).
We can use the equation of motion:
h = ut + (1/2)gt²
Where:
h = height (6.9 m)
u = initial velocity (0 m/s)
g = acceleration due to gravity (9.8 m/s²)
t = time (what we want to find)
Plugging in the values into the equation, we have:
6.9 = (0)t + (1/2)(9.8)t²
Simplifying the equation gives:
6.9 = 4.9t²
Rearranging the equation, we have:
t² = 6.9 / 4.9
t² ≈ 1.408
Taking the square root of both sides, we get:
t ≈ √1.408
t ≈ 1.19 seconds
Therefore, the water-filled balloon will be in the air approximately 1.19 seconds before reaching the ground.