A flowerpot falls from a windowsill 25.0 m above the sidewalk.
A. How fast is the flowerpot moving when it strikes the ground??
B. How much time does a cat on the sidewalk below have to move out of the way before the flowerpot hits the ground?
(Please answer with step by step solutions)
To solve these problems, we can use the equations of motion under constant acceleration:
1. Finding the speed of the flowerpot before impact:
The initial velocity of the flowerpot, u = 0 (it starts from rest).
The acceleration due to gravity, g = -9.8 m/s² (negative because it acts downward).
The displacement, s = -25.0 m (negative because it is downward).
Using the equation v² = u² + 2as, we can find the final velocity, v.
v² = u² + 2as
v² = 0 + 2(-9.8)(-25.0)
v² = 490
v = √490 ≈ 22.1 m/s
A. The flowerpot is moving at approximately 22.1 m/s just before it strikes the ground.
2. Finding the time a cat has to move out of the way:
We need to find the time it takes for the flowerpot to fall from 25.0 m.
Using the equation s = ut + (1/2)at², we can find the time, t.
s = ut + (1/2)at²
-25.0 = 0 + (1/2)(-9.8)t²
-25.0 = -4.9t²
t² = 25.0 / 4.9
t ≈ √(25.0 / 4.9)
t ≈ 2.03 s
B. The cat has approximately 2.03 seconds to move out of the way before the flowerpot hits the ground.
Note: In these calculations, we have assumed there is no air resistance acting on the flowerpot.
To get the answer to these questions, we can use the principles of kinematics and Newton's laws of motion. We'll start by solving part A of the question.
A. How fast is the flowerpot moving when it strikes the ground?
To find the velocity of the flowerpot just before it hits the ground, we need to use the equation of motion:
v^2 = u^2 + 2as
Here,
v = final velocity (which we want to find)
u = initial velocity (0 m/s as the pot starts from rest)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward as negative)
s = displacement (25.0 m, since it falls from a height of 25.0 m)
Substituting these values into the equation, we get:
v^2 = (0 m/s)^2 + 2 * (-9.8 m/s^2) * (25.0 m)
v^2 = -490 m^2/s^2
Note: The negative sign indicates the direction (downward).
To get the magnitude of the velocity (speed), we take the square root of both sides:
v = √(-490 m^2/s^2)
v = 22.1 m/s (rounded to one decimal place)
So, the flowerpot is moving at a speed of 22.1 m/s just before it strikes the ground.
Now, let's move on to part B.
B. How much time does a cat on the sidewalk below have to move out of the way before the flowerpot hits the ground?
To find the time it takes for the flowerpot to hit the ground, we can use the equation of motion:
s = ut + (1/2)at^2
Here,
s = displacement (25.0 m)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (which we want to find)
Substituting the values into the equation, we have:
25.0 m = (0 m/s)(t) + (1/2)(-9.8 m/s^2)(t^2)
Simplifying, we get:
25.0 m = (1/2)(-9.8 m/s^2)(t^2)
Dividing both sides by (1/2)(-9.8 m/s^2), we obtain:
t^2 = (25.0 m) / ((1/2)(-9.8 m/s^2))
t^2 = 5.102 m / s^2
Taking the square root of both sides, we find:
t = √(5.102 m / s^2)
t = 2.26 s (rounded to two decimal places)
Therefore, the cat on the sidewalk has approximately 2.26 seconds to move out of the way before the flowerpot hits the ground.
IONS
A. The object follows uniformly accelerated motion. We can use the formula,
vf^2 - vo^2 = 2gd
where
vf = final or terminal velocity, m/s
vo = initial velocity, m/s
g = acceleration due to gravity = 9.8 m/s^2
Since the object fell, its initial velocity is zero. Substituting,
vf^2 - 0 = 2*9.8*25
vf^2 = 490
vf = ?
B. We can solve for the time before it hits the ground using this formula:
h = vo*t - (1/2)gt^2
Substituting,
-25 = 0 - (0.5)(9.8)(t^2)
-25 = -4.9t^2
5.102 = t^2
t = ?
Hope this helps~ `u`
B.