Question

A rocket is launched into the air and follows the path, h(t) = -3t^2 + 12t, where t is time measured in seconds. Determine how long it takes the rocket to hit the ground. Show all work for full credit.

Answer 1

note that height is equal to 0 when time equals 0 and when time equals 4, but we are only concerned with time equal to 4

Answer 2

Answer:

4 seconds

Step-by-step explanation:

Given: A rocket is launched into the air and follows the path,$\text{h}(\text{t})=-3\text{t}^2+12\text{t}$, where t is time measured in seconds

To Find: how long it takes the rocket to hit the ground

Solution:

Path followed by rocket

$\text{h}(\text{t})=-3\text{t}^2+12\text{t}$

here, t is time in seconds and h is height

height of rocket  when it hit ground,

$\text{h}=0\text{m}$

putting value of h in equation

$0=-3\text{t}^2+12\text{t}$

$3\text{t}^2-12\text{t}=0$

$3\text{t}(\text{t}-4)=0$

$\text{t}=0,4$

height of rocket will be 0 at time t=0s and t=4s

$\text{t}=0$ , time when rocket is launched

$\text{t}=4$ , time when rocket hits ground

Total time taken to hit the ground$=4-0=4\text{s}$

Time taken by rocket to hit the ground is 4 seconds.