Question

A model rocket is launched straight upward. The path of the rocket is modeled by h = -16t2 + 200t, where h represents the height of the rocket and t represents the time in seconds. a. What is its maximum height? b. Is it still in the air after 8 seconds? Explain why or why not. c. Is it still in the air after 14 seconds? Explain why or why not.

Answer 1

Answer:

a) 625

b) Yes

c) No

Step-by-step explanation:

a)To calculate the maximum height we can use the derivative to find the slope of the function.

$dh/dt=-32t+200$

To find the slope we make the equation equal to 0:

$-32t+200=0$

$t=200/32$

The slope at 0 is t=200/32=6.25[/tex]

The maximum height at t=6.25 is

$h=-16(6.25^2)+200(6.25)=625$

The maximum height is 625

b) at t=8

$-16(8^2)+200(8)=576$

The height will be 576 which is less than 625

The two x-intercepts are when h=0:

c)

$-16(14^2)+200(14)=-336$

$-16t^2+200t=0$

$-2t^2+25t=0$

$t(-2t+25)=0$

$t=0$

$t=12.5$

The ball is not in the air because the ball will only be in the air from t=0 to t=12.5