Question

a rocket is launched vertically upward from the ground with an initial velocity of 12 ft/s. the height of the rocket in t seconds is given by s(t)=-16t² + 12t. determine how long the rocket is in the air. enter an exact answer.
answer below:

Explanation:

Step1: Set height function to 0

When the rocket hits the ground, $s(t)=0$. So we set $-16t^{2}+12t = 0$.

Step2: Factor out t

$t(-16t + 12)=0$.

Step3: Solve for t

We have two solutions from the zero - product property: $t = 0$ or $-16t+12=0$. Solving $-16t + 12=0$ gives $16t=12$, so $t=\frac{12}{16}=\frac{3}{4}$. The $t = 0$ solution corresponds to the time of launch, and $\frac{3}{4}$ seconds is the time when the rocket returns to the ground.

Answer:

$\frac{3}{4}$