a brick is dropped form the roof of a building. On the way down it passes a 2m high window and is observed to pass from the top to the bottom of the window in 0.25s. (a) How fast was it moving when it passed the top of the window? (b)How far below the roof of the building is the top of the window?

average speed passing window

= -2 meters/.25 s = -8 ms

constant acceleration = -9.81 m/s
so while crossing the window
v = Vi -.981t
change in v while crossing window = -9.81*.25 = -2.4525 m/s
half that = -1.23 m/s
so v at window top
= -8+1.23 = -6.77 m/s

so how far did it fall from the roof to the window top?
v = 0 -9.81 t
-6.77 = -9.81 t
t = .69 seconds
d = .49 t^2
=.233 meters

TY very much Damon

To solve this problem, we can use the equations of motion and apply them to the given scenario.

(a) How fast was the brick moving when it passed the top of the window?

To find the velocity of the brick when it passed the top of the window, we can use the equation of motion:

v = u + at,

where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time taken.

In this case, we know the final velocity is 0 (since the brick is at rest at the bottom of the window), the time taken is 0.25 seconds, and we need to find the initial velocity when it passed the top of the window.

Let's denote the initial velocity as v₀.

Using the equation of motion, we have:

0 = v₀ + at.

Since the brick is in free fall, its acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s².

Substituting the known values into the equation, we have:

0 = v₀ + (9.8 m/s²)(0.25 s).

Simplifying the equation, we get:

v₀ = -(9.8 m/s²)(0.25 s).

Calculating the value, we find:

v₀ = -2.45 m/s.

Therefore, the brick was moving at a speed of approximately 2.45 m/s downwards when it passed the top of the window.

(b) How far below the roof of the building is the top of the window?

To find the distance below the roof of the building where the top of the window is located, we can use the equation of motion:

s = ut + (1/2)at²,

where:
s is the displacement,
u is the initial velocity,
t is the time taken, and
a is the acceleration.

In this case, we know the initial velocity is v₀ (which we found to be -2.45 m/s), the time taken is 0.25 seconds, and we need to find the displacement.

Let's denote the displacement as s.

Using the equation of motion, we have:

s = (v₀)(t) + (1/2)at².

Substituting the known values into the equation, we have:

s = (-2.45 m/s)(0.25 s) + (1/2)(9.8 m/s²)(0.25 s)².

Simplifying the equation, we get:

s = -0.6125 m + 0.030625 m.

Calculating the value, we find:

s = -0.581875 m.

Therefore, the top of the window is approximately 0.581875 meters below the roof of the building.

last two lines

d = 4.9 t^2
= 2.33 meters