Your on a roller coaster when a shoe falls off. The function y=300−16t^2 represents the height y of the shoe t seconds after it falls

Question

y is in feet  The shoe lands on the top of an 11-foot-tall building. After how many seconds does the shoe hit the building?

Answer 1

Arturo has a valid point, however, it is not unusual for text books

to over simplify problems of this sort.  The height of a free falling

object over time is given by:

h(t) = (-1/2)gt² + v₀t + h₀

g = acceleration of gravity = 32 ft/sec²

v₀ = initial velocity = 0 for this problem as stated

h₀ = initial height = 300 ft for this problem

 

h(t) = -16t² + 300

We want to solve for t when h(t) = 11

 

-16t² + 300 = 11

-16t² + 289 = 0

This is actually the difference of two squares

289 - 16t² = 17² - (4t)² = (17-4t)(17+4t)

 

So we have:  (17-4t)(17+4t) = 0

Either: 17-4t = 0

           4t = 17

           t = 17/4

           t = 4 1/4 seconds

 

OR:   17+4t = 0

         4t = -17

         t = -4 1/4 seconds

 

Discarding the negative, it will take 4 1/4 or 4.25 seconds to hit the building