A flowerpot falls from a window sill 25.3 m

above the sidewalk.
What is the velocity of the flowerpot when
it strikes the ground? The acceleration of
gravity is 9.81 m/s
2
.
Answer in units of m/s
How much time does a passerby on the sidewalk below have to move out of the way before
the flowerpot hits the ground?
Answer in units of s

To find the velocity of the flowerpot when it strikes the ground, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity (which is 0 as the flowerpot falls)
a = acceleration due to gravity (which is 9.81 m/s^2)
s = distance fallen (25.3 m)

We can rearrange the equation to solve for v:

v^2 = 0^2 + 2 * 9.81 * 25.3
v^2 = 0 + 495.2466
v = sqrt(495.2466)
v ≈ 22.254 m/s

So, the velocity of the flowerpot when it strikes the ground is approximately 22.254 m/s.

To find the time a passerby on the sidewalk below has to move out of the way before the flowerpot hits the ground, we can use another equation of motion:

s = ut + (1/2)at^2

Where:
s = distance fallen (25.3 m)
u = initial velocity (0 m/s)
a = acceleration due to gravity (9.81 m/s^2)
t = time

We can rearrange the equation to solve for t:

25.3 = 0 * t + (1/2) * 9.81 * t^2
25.3 = 4.905 * t^2
t^2 = 25.3 / 4.905
t ≈ sqrt(25.3 / 4.905)
t ≈ sqrt(5.1602)
t ≈ 2.273 s

So, a passerby on the sidewalk below has approximately 2.273 seconds to move out of the way before the flowerpot hits the ground.