Derivatives

Question

A man on the 12th floor of a building sees a bucket (dropped by a window washer) pass his window and notes that it hits the ground 1.3 s later. Assuming a floor is 12 ft high (and neglecting air friction), from which floor was the bucket dropped?

Answer 1

Break this problem into parts.

 

(1)

 

Find the speed at the 12th floor, which is at 144 ft.

 

It takes time t to reach the ground (height zero) from the 12th floor, and t = 1.3 s.

 

0 = 144 - vt - 16t² = 144 - 1.3v - 16(1.3)²

 

v ≅ 89.97 ft/s in the down direction

 

(2)

 

Use v from part (1) to get the time t from the top at h₀ to 144 ft, where v at 144 ft feet was found to be about 89.97 ft/s.  Assuming it was released from rest,

 

v(t) = gt = 32t

 

89.97 = 32t

 

t = 89.97/32 s ≅ 2.812 s

 

Again, assuming it was released from rest,

 

h(t) = -16t² + h₀

 

h = 144

t = 2.812 s

h₀ = initial height

 

144 = -16(2.812)² + h₀

 

h₀ ≅ 270.5 ft

 

Divide h₀ by 12 to get the floor.  It looks like it was dropped between floors.

 

Here is a way to test the answer:  Evaluate h(t) when t = sum of time from top to 144 ft and time from 144 ft to bottom.  It should give zero, allowing for roundoff.

 

t = 1.3 + 2.812 = 4.112

 

h = h₀ - 16t² = 270.5 - 16(4.112)² ≅ -0.0367

 

(Close enough to zero, allowing for roundoff. )