Answered

Join IDNLearn.com today and start getting the answers you've been searching for. Get comprehensive and trustworthy answers to all your questions from our knowledgeable community members.

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 :

Sqdancefanidn

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.

We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.