Bill kicks a rock off the top of his apartment building. It strikes a window across the street 15m away. The acceleration of gravity is 9.8 m/s^2. If the window is 13m below the position where Bill contacted the rock, how long was it in the air?

9.14

To find the time the rock was in the air, we can use the equation of motion for vertical displacement:

Δy = v₀t + (1/2)gt²

Where:
Δy = vertical displacement (height difference between the initial position and the final position)
v₀ = initial vertical velocity (which is 0 since Bill drops the rock)
t = time
g = acceleration due to gravity

Given:
Δy = 13 m
g = 9.8 m/s²

Rearranging the equation, we have:

0 = (1/2)gt² + v₀t - Δy

Since the rock is dropped, its initial vertical velocity is 0. Substituting v₀ = 0:

0 = (1/2)gt² - Δy

Simplifying further, we get:

(1/2)gt² = Δy

Now, plug in the values:

(1/2)(9.8 m/s²)t² = 13 m

Rearrange the equation:

4.9t² = 13

Divide both sides by 4.9:

t² = 13 / 4.9

t² ≈ 2.653

Taking the square root of both sides:

t ≈ √2.653

t ≈ 1.63 seconds

Therefore, the rock was in the air for approximately 1.63 seconds.