A flowerpot falls from a windowsill 33.0 m above the sidewalk.
a) How much times does a passerby on the sidewalk below have to move out of the way before the flowerpot
hits the ground?
b) What is the flowerpot's velocity when it strikes the ground?
g = a = -9.81 m/s^2
v = 0 - 9.81 t
h = 33.0 + 0 t - 4.9 t^2
h = 0 at crash
4.9 t^2 = 33
t = 2.6 seconds to disaster
back to v = 0 -9.81 t
= - 9.81*2.6 , negative because downward like g
To find the answers to these questions, we can use the equations of motion in physics. Specifically, we can use equations of motion related to free fall, since the flowerpot falls under the force of gravity.
a) To calculate how much time the passerby on the sidewalk has to move out of the way, we can use the following equation:
d = 1/2 * g * t^2
Where:
d is the vertical distance traveled by the flowerpot (33.0 m in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken
Rearranging the equation, we get:
t = sqrt(2d / g)
Substituting the values, we have:
t = sqrt(2 * 33.0 / 9.8)
t ≈ 2.86 seconds
Therefore, the passerby has approximately 2.86 seconds to move out of the way before the flowerpot hits the ground.
b) To calculate the velocity of the flowerpot when it strikes the ground, we can use the equation:
v = sqrt(2g * d)
Substituting the values:
v = sqrt(2 * 9.8 * 33.0)
v ≈ 23.1 m/s
Therefore, the flowerpot's velocity when it strikes the ground is approximately 23.1 m/s.