Bill kicks a rock off the top of his apartment building. It strikes a window across the street 18m away. The acceleration of gravity is 9.8 m/s sqaured.

If the window is 12 m below the position where bill contacted the rock, how long was it in the air?
Answer in units of s.

38.5

Hegehdu

To find the time it took for the rock to travel from the top of the building to the window, we can use the equation of motion:

d = vit + (1/2)at^2

In this case, the initial vertical velocity (vi) is 0 m/s because the rock was kicked vertically downwards. The distance (d) is 12 m (the height of the window below the starting point), and the acceleration (a) is the acceleration due to gravity, which is -9.8 m/s^2 (negative because it acts downwards).

Plugging in these values into the equation, we have:

12 = (1/2)(-9.8)t^2

Simplifying further, we get:

24 = -4.9t^2

Dividing both sides by -4.9, we have:

t^2 = -24 / -4.9

t^2 ≈ 4.898

Taking the square root of both sides, we get:

t ≈ √4.898

t ≈ 2.21 s

Therefore, it took approximately 2.21 seconds for the rock to travel from the top of the building to the window.

To find the time the rock was in the air, we can use the kinematic equation:

h = vit + (1/2)at^2

Where:
h = height (in this case, 12m)
vi = initial velocity (which is 0 since the rock starts from rest)
a = acceleration due to gravity (-9.8 m/s^2, taking into account the direction)
t = time

We want to solve for t. Rearranging the equation:

h = (1/2)at^2
2h = at^2
t^2 = (2h) / a
t = √(2h / a)

Substituting the given values:
t = √(2 * 12 m / 9.8 m/s^2)

Calculating this:

t ≈ √(24 / 9.8)
t ≈ √2.449
t ≈ 1.57 s

Therefore, the rock was in the air for approximately 1.57 seconds.