Bill kicks a rock off the top of his apartment
building. It strikes a window across the street
11 m away.
The acceleration of gravity is 9.8 m/s
2
.
If the window is 22 m below the position
where Bill contacted the rock, how long was
it in the air?
Answer in units of s
do this one just like the one with the cat.
To find the time the rock was in the air, we can use the kinematic equation for vertical motion under constant acceleration:
h = v_initial * t + (1/2) * a * t^2
where:
h is the vertical displacement (distance fallen by the rock),
v_initial is the initial velocity (0 m/s since it was kicked off only by Bill),
a is the acceleration (gravity, -9.8 m/s^2 since it is acting downward),
t is the time the rock was in the air.
Given that the rock fell a distance of 22 m, we can rearrange the equation to solve for t:
h = (1/2) * a * t^2
22 = (1/2) * (-9.8) * t^2
22 = -4.9t^2
Next, we can solve for t^2 by dividing both sides by -4.9:
t^2 = 22 / -4.9
Taking the square root of both sides gives us:
t = √(22 / -4.9)
Calculating this value (considering the negative value for time is non-physical), we get:
t ≈ 1.45 seconds
Therefore, the time the rock was in the air is approximately 1.45 seconds.