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?
Whoops, the question is: A flowerpot falls from a window sill 37.4 m above the sidewalk.
h= 1/2 g t^2 solve for h.
To find the velocity of the flowerpot when it strikes the ground, we need to use the equations of motion.
A) The first step is to determine the time it takes for the flowerpot to fall to the ground. We can do this by using the equation of motion:
S = ut + (1/2)at^2
where:
S is the distance (height) the flowerpot falls,
u is the initial velocity (0 m/s since the flowerpot was dropped),
a is the acceleration due to gravity (-9.81 m/s²),
and t is the time it takes for the flowerpot to fall.
Since the flowerpot is falling from rest, the initial velocity is 0 m/s. Rearranging the equation, we get:
S = (1/2)at^2
Now, we know the acceleration due to gravity is -9.81 m/s² and let's assume the height of the building is given as 20 meters. Plugging these values in, we get:
20 = (1/2)(-9.81)t^2
Simplifying the equation further:
t^2 = (20 * 2) / 9.81
t^2 = 40 / 9.81
t ≈ 2.02 seconds
So, it takes approximately 2.02 seconds for the flowerpot to strike the ground.
Next, to find the velocity of the flowerpot when it strikes the ground, we use the equation:
v = u + at
Since the initial velocity, u, is 0 m/s, the equation simplifies to:
v = at
Plugging in the acceleration due to gravity (-9.81 m/s²) and the time (2.02 seconds) obtained earlier:
v = (-9.81 m/s²)(2.02 s)
v ≈ -19.8 m/s
Therefore, the velocity of the flowerpot when it strikes the ground would be approximately -19.8 m/s (negative sign indicates downward direction).
B) To calculate the time a passerby on the sidewalk has to move out of the way before the flowerpot hits the ground, we need to consider the distance the passerby needs to cover horizontally.
Again, using the equation of motion:
S = ut + (1/2)at^2
Where:
S is the horizontal distance the passerby needs to cover,
u is the initial horizontal velocity (0 m/s since the pedestrian is at rest),
a is the horizontal acceleration (0 m/s² since the pedestrian is not subjected to any horizontal force),
and t is the time the passerby has to move out of the way.
Since there is no horizontal acceleration, the distance S covered by the passerby is directly dependent on the horizontal velocity, which can be calculated using the equation:
S = vt
Plugging in the velocity (v) obtained earlier (-19.8 m/s) and rearranging the equation, we have:
t = S / v
Assuming the passerby wants to move out of the way before the pot hits the ground at a horizontal distance of 5 meters, we have:
t = 5 m / (-19.8 m/s)
t ≈ -0.25 s
The negative sign indicates that the passerby needs to move out of the way 0.25 seconds before the pot hits the ground.