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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.

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