Calculus, derivatives

Question

A man on the 12th floor of a building sees a bucket passing his window and notes that it hits the ground 1 second later. Assuming the floor is 4.9metres high. From what floor was bucket dropped. g = 9.8

Answer 1

The 12th floor is 58.8 meters high.

We can use the falling distance formula:

h(t) = gt²/2 + vt + h = -4.9t²+vt + 58.8

After one second, h(t) = 0, so we can solve for the velocity when passing the 12th floor.

-4.9 + v +58.8 = 0

v = -53.9 m/s

Using v = gt we can determine the time it took to reach the 12th floor starting with v=0

t = -53.9/-9.8 = 5.5 seconds

h = gt²/2 = 4.9(5.5)² = 148.225 meters or 148.225/4.9 floors = 30.25 floors ≈ 31 floors

31+12 = 43rd floor

Answer 2

there is some ambiguity as to where the 12th floor is

ground level the 0th floor is 0 meters or is that the basement? from -4.9 to 0?

if the 1st floor is from 0 to 4.9 meters high

2nd is from 4.9 to 9.8

nth floor is from (n-1)4.9 to (n-1)4.9+ 4.9

12th floor is from (11)4.9 to (11)4.9+ 4.9 = 53.9 to 58.8 meters high

was the bucket dropped from the bottom, top or somewhere in between 53.9 to 58.8?

probably about a window's height = about 1 or 2 meters from the bottom of the 12th floor or from about 55 to 56 meters

the bucket was dropped from the 42.4th floor window which was 41.4(4.9) to 41.4(4.9)+4.9 = about 202 meters high

a little less than the midpoint of the 42nd floor

on the 12th floor the bucket likely passed by a similar window about the 12.4th floor, when it had 1 second left to reach the ground or 0th floor

h(t) = -4.9t^2 + ho where ho = initial height = 202 = 41.4(4.9) = about (6.44)^2(4.9)

velocity at time t = v(t) = h'(t) = -4.9(2)t

in 6.44 seconds v(6.44) = -4.9(2)(6.44) = -4.9(12.88) at the 12.4th floor

v(0) = zero velocity initially, on the 41.4th floor

v(6.44) = -4.9(2)(6.44) = -4.9(12.88) after almost 6 1/2 seconds

v(7) =-4.9(2)(6.44)^2 on the 41.44-41.47th floor = 0.03 meter below ground level