A flowerpot falls from a window sill 37.6 m above the sidewalk.?
A) What is the velocity of the flowerpot when it strikes the ground? The acceleration of gravity is 9.81 m/s2 ?
B) How much time does a passerby on the side-walk below have to move out of the way before the flowerpot hits the ground?
655.308
try
To calculate 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 (in this case, the flowerpot is initially at rest, so u = 0)
a = acceleration (which is gravity, -9.81 m/s^2 in this case)
s = displacement (the distance the flowerpot falls, which is 37.6 m)
By substituting the given values into the equation and solving for v, we can find the velocity when the flowerpot strikes the ground.
A) Calculating the velocity of the flowerpot:
v^2 = 0^2 + 2(-9.81)(37.6)
v^2 = 0 - 2(9.81)(37.6)
v^2 = -2(9.81)(37.6)
v^2 = -733.6
v = √(-733.6)
Since the square root of a negative number is not real, it means that the flowerpot does not reach the ground with a positive velocity. In other words, the flowerpot hits the ground with a velocity of 0 m/s.
B) Calculating the time a passerby on the side-walk has to move out of the way:
To find the time it takes for the flowerpot to hit the ground, we can use the equation of motion:
s = ut + 0.5at^2
Where:
s = displacement (the distance the flowerpot falls, which is 37.6 m)
u = initial velocity (0 m/s)
a = acceleration (gravity, -9.81 m/s^2)
t = time
By substituting the given values into the equation and solving for t, we can find the time a passerby has to move out of the way before the flowerpot hits the ground.
37.6 = 0*t + 0.5*(-9.81)*t^2
Rearranging the equation:
18.9*t^2 = 37.6
t^2 = 37.6/18.9
t^2 = 2
Taking the square root of both sides:
t = √2
The time it takes for the flowerpot to hit the ground is approximately 1.41 seconds. Therefore, a passerby on the side-walk below has around 1.41 seconds to move out of the way before the flowerpot hits the ground.