Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 50 % of the entire distance it falls. How tall is the building?

Sagot :

sqdancefan

Answer:

  186.5 ft

Explanation:

The height (in feet) as a function of time for a building height of 'b' is ...

  h(t) = -16t^2 +b

This will be zero for some time T.

  h(T) = 0 = -16T^2 +b

and it will be b/2 for T-1:

  b/2 = -16(T-1)^2 +b

Solving each equation for b gives ...

  16T^2 = b = 32(T-1)^2

  0 = 16T^2 -64T +32

  (T -2)^2 -2 = 0 . . . . . . . . divide by 16, rearrange to vertex form

  T = 2 +√2 . . . . . . . . . . add 2, square root, add 2

  b = 16T^2 = 16(6 +4√2) ≈ 186.510 . . . . feet

The building is about 186.5 feet high.

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.