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)

$\frac{h(3)-h(0)}{3}=\\ \frac{300-16*3^2-(300-16*0^2)}{3}=\\ \frac{300-16*3^2-(300)}{3}=\\ \frac{-16*3^2}{3}=\\ -16*3=-48$

Answer 2

Answer:

-96 feet/sec.

Step-by-step 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 $\frac{d[h(t)]}{dt}=\frac{d(300-16t)^{2} }{dt}$

Speed = $\frac{d}{dt}(300)$ - $\frac{d}{dt}(16t^{2} )$

         = -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.