Question

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the function h(t)=300-16t2. Which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall.

Answer 1

your average speed is the average difference over time (in this case 3 seconds)


`(h(3)-h(0))/(3)=\\ (300-16*3^2-(300-16*0^2))/(3)=\\ (300-16*3^2-(300))/(3)=\\ (-16*3^2)/(3)=\\ -16*3=-48`

Answer 2

-96 feet/sec.

Explanation:

The height h of a falling object after t second when it is dropped from a platform 300 feet above the ground is modeled by the function h (t) = 300 - 16t²

Then speed or average rate by which the object falls from height h will be
`(d[h(t)])/(dt)=(d(300-16t)^(2) )/(dt)`

Speed =
`(d)/(dt)(300)` -
`(d)/(dt)(16t^(2) )`

= -16 (2t)

Speed = -32t

After 3 second speed of object will be

Speed = -32(3)

= -96 feet/second

Therefore, the average rate will be -96 feet/sec. during the first 3 seconds of its fall.