If a stone falls past a bird on a ledge moving at 4 m/s, how fast will the stone be moving just before it hits the ground below 9 seconds later?
To determine the speed of the stone just before it hits the ground, we need to use the equations of motion. Specifically, we can use the kinematic equation:
v = u + at
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
In this case, the stone is affected by the acceleration due to gravity, which is approximately 9.8 m/s^2 (neglecting air resistance). Since the stone is initially at rest, its initial velocity, u, is 0 m/s.
We can rearrange the equation to solve for v:
v = u + at
v = 0 + (9.8 m/s^2) * (9 s)
Substituting the values:
v = 0 + (9.8 m/s^2) * (9 s)
v = 88.2 m/s
Therefore, the stone will be moving at a speed of approximately 88.2 m/s just before it hits the ground, assuming no air resistance.