Bill kicks a rock off the top of his apartment building. It strikes a window across the street 12 m away. The acceleration of gravity is 9.8 m/s^2. If the window is 16 m below the position where Bill contacted the rock, how long was it in the air?
h = Vo*t + 0.5g*t^2 = 16
0 + 4.9t^2 = 16
t^2 = 3.27 s.
To find the time the rock was in the air, we can use the kinematic equation:
y = yi + vi*t + (1/2)*a*t^2
Where:
- y is the final vertical position of the rock (16 m below the starting position)
- yi is the initial vertical position of the rock (from the starting position, which is 0 m)
- vi is the initial vertical velocity of the rock (0 m/s since the rock is initially at rest)
- t is the time in the air
- a is the acceleration due to gravity (-9.8 m/s^2 since it acts downward)
Plugging in the known values:
16 = 0 + 0*t - (1/2)*9.8*t^2
Simplifying the equation:
16 = -4.9*t^2
Rearranging the equation to isolate t^2:
t^2 = 16 / (-4.9)
t^2 = -3.2653
Since time cannot be negative, we can disregard the negative solution.
Taking the square root of t^2:
t = sqrt(3.2653)
t ≈ 1.81 seconds
Therefore, the rock was in the air for approximately 1.81 seconds.