A flowerpot falls from a window sill 26.8 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 .
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*26.8 = 525.28
V = 22.92 m/s.
To find the velocity of the flowerpot when it strikes the ground, you can use the equation of motion for an object in free fall:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (which is 0 because the flowerpot starts from rest)
a = acceleration due to gravity (given as 9.81 m/s^2)
s = displacement (26.8 m, since the flowerpot falls from a height of 26.8 m)
Plugging in the values we have:
v^2 = 0^2 + 2 * 9.81 * 26.8
Simplifying:
v^2 = 2 * 9.81 * 26.8
v^2 = 527.4256
Taking the square root of both sides:
v ≈ √527.4256
v ≈ 22.98 m/s
Therefore, the velocity of the flowerpot when it strikes the ground is approximately 22.98 m/s.