A flowerpot falls from a window sill 38.8 m
above the sidewalk.
What is the velocity of the flowerpot when
it strikes the ground? The acceleration of
gravity is 9.81 m/s2 .
Answer in units of m/s.
How much time does a passerby on the side-
walk below have to move out of the way before
the flowerpot hits the ground?
Answer in units of s.
taylor's window is 6 feet off the ground and the sidewalk is 25 feet from his house
To find the velocity of the flowerpot when it strikes the ground, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (what we want to find)
u = initial velocity (0 m/s, as the pot starts from rest)
a = acceleration due to gravity (-9.81 m/s^2, since it acts in the opposite direction of motion)
s = displacement (38.8 m, distance fallen)
Let's substitute these values into the equation and solve for v:
v^2 = 0 + 2(-9.81)(38.8)
v^2 = -2(9.81)(38.8)
v^2 = -2(379.128)
v^2 = -758.256
v ≈ √(-758.256)
Since the acceleration due to gravity is acting in the opposite direction of motion, the velocity will be negative. However, in this case, we are only interested in the magnitude of the velocity, so we can disregard the negative sign:
v ≈ √758.256
v ≈ 27.55 m/s
Therefore, the velocity of the flowerpot when it strikes the ground is approximately 27.55 m/s.
To calculate the time a passerby on the sidewalk has to move out of the way, we can use another kinematic equation:
s = ut + (1/2)at^2
Where:
s = displacement (38.8 m, distance fallen)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.81 m/s^2)
t = time (what we want to find)
Let's substitute these values into the equation and solve for t:
38.8 = 0 + (1/2)(-9.81)t^2
38.8 = -4.905t^2
t^2 = -38.8 / -4.905
t^2 = 7.92
t ≈ √7.92
We can ignore the negative square root since time cannot be negative:
t ≈ √7.92
t ≈ 2.81 s
Therefore, a passerby on the sidewalk below has approximately 2.81 seconds to move out of the way before the flowerpot hits the ground.