1. A flowerpot falls from a windowsill 25.0 m above the sidewalk.
a. how fast is the flowerpot moving when it strikes the ground?
dont i need to know the time it takes, what formula am i supposed to use for this? would 25 m be the displacement?
b. how much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground?
please help!!
wewt
To answer part a) of your question, we can use the kinematic equation to determine the final velocity of the flowerpot when it strikes the ground.
The equation we can use is:
v^2 = u^2 + 2as
where:
v = final velocity (which we want to find)
u = initial velocity (which is 0, since the flowerpot starts from rest)
a = acceleration due to gravity (which is approximately 9.8 m/s^2, assuming no air resistance)
s = displacement (which is 25.0 m, as you correctly mentioned)
Plugging in the values, we have:
v^2 = 0^2 + 2(9.8)(25.0)
v^2 = 490
v = √490
v ≈ 22.1 m/s
Therefore, the flowerpot is moving at approximately 22.1 m/s when it strikes the ground.
Now, for part b) of your question, we can use another kinematic equation to find the time it takes for the flowerpot to fall from the windowsill to the ground.
The equation we can use is:
s = ut + 1/2at^2
Since the initial velocity (u) is 0 m/s and the displacement (s) is 25.0 m, we can rearrange the equation to solve for time (t):
25.0 = 0*t + 1/2*(9.8)*t^2
25.0 = 4.9*t^2
t^2 = 25.0/4.9
t ≈ √(25.0/4.9)
t ≈ 2.04 s
Therefore, a passerby on the sidewalk below has approximately 2.04 seconds to move out of the way before the flowerpot hits the ground.
I hope this helps! Let me know if you have any further questions.
To answer these questions, we can use the equations of motion. Let's break it down step by step:
a. To find the speed of the flowerpot when it strikes the ground, we need to calculate the final velocity. We can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s, as the flowerpot starts from rest)
a = acceleration due to gravity (-9.8 m/s^2, as it acts downward)
s = displacement (25.0 m, as the flowerpot falls 25.0 m)
Plugging in these values into the equation, we get:
v^2 = 0^2 + 2(-9.8)(25.0)
v^2 = -490
(v is negative because it represents the direction of motion, opposite to gravity)
To find the velocity (v), we can take the square root of both sides of the equation:
v = √(-490)
(Note: The square root of a negative value is an imaginary number, -490 in this case)
Since we are interested in the magnitude of the velocity (speed), the answer is the absolute value of this imaginary number:
Speed = |v| = |-22.14| = 22.14 m/s
Therefore, the flowerpot is moving at a speed of 22.14 m/s when it strikes the ground.
b. To find the time a passerby has to move out of the way before the flowerpot hits the ground, we need to calculate the time taken (t). We can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement (25.0 m)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)
Plugging in these values into the equation, we get:
25 = 0t + (1/2)(-9.8)(t^2)
25 = -4.9t^2
Rearranging the equation, we have:
4.9t^2 = -25
Dividing both sides by 4.9, we get:
t^2 = -25 / 4.9
Again, we encounter a square root of a negative value, which leads to an imaginary number. Since time cannot be imaginary, it means that there is no real solution for t. Therefore, a passerby on the sidewalk below does not have any time to move out of the way as the flowerpot falls directly to the ground.
Note: In reality, a passerby would not be able to dodge a falling object like a flowerpot at the last moment, so it's crucial to be cautious and avoid walking beneath open windows or unsafe areas.
(a)
Vf^2 = Vo^2 + 2gy
since Vo = 0,
Vf^2 = 2gy
Substitute and solve for Vf
(b)
y =(1/2) gt^2
Substitute and solve for t