In the problem, please assume the free-fall acceleration g = 9.80 m/s2 unless a more precise value is given in the problem statement. Ignore air resistance.

A stone is thrown vertically downward from the roof of a building. It passes a window 15.0 m below the roof with a speed of 24.3 m/s. It lands on the ground 4.75 s after it was thrown.

is there a question in there somewhere?

(a) What was the initial velocity of the stone?

(b) How tall is the building?

sorry I added the question to another post, but here it is. Thanks :)

To solve this problem, we can use the equations of motion for free-fall to determine the height of the building.

Given:
g = 9.80 m/s² (acceleration due to gravity)
h = 15.0 m (height of the window)
v = 24.3 m/s (velocity of the stone)
t = 4.75 s (time taken to reach the ground)

First, we can determine the initial velocity (u) of the stone using the formula v = u + gt, where v is the final velocity, t is the time, and g is the acceleration due to gravity. Plugging in the values we have:

24.3 m/s = u + (9.80 m/s²)(4.75 s)

Now, let's solve for u:

u = 24.3 m/s - (9.80 m/s²)(4.75 s)

u = 24.3 m/s - 46.55 m/s

u = -22.25 m/s

The negative sign indicates that the initial velocity of the stone is downwards.

Next, we can use the kinematic equation h = ut + (1/2)gt² to find the height (h) of the building. Plugging in the known values:

15.0 m = (-22.25 m/s)(4.75 s) + (0.5)(9.80 m/s²)(4.75 s)²

Simplifying and solving for h:

15.0 m = -105.69 m + (0.5)(9.80 m/s²)(22.5625 s²)

15.0 m = -105.69 m + (111.0625 m)

15.0 m = 5.3725 m

Therefore, the height of the building is 5.3725 meters.