Question

a rocket is launched in the air. its height in feet is given by h(t)= -16t² + 144t where t represents the time in seconds after launch. after how many seconds does the rocket hit the ground?

Explanation:

Step1: Set height equal to 0

When the rocket hits the ground, $h(t)=0$. So we set $-16t^{2}+144t = 0$.

Step2: Factor out common term

Factor out $-16t$ from the left - hand side of the equation: $-16t(t - 9)=0$.

Step3: Use zero - product property

If $ab = 0$, then either $a = 0$ or $b = 0$. So, $-16t=0$ or $t - 9=0$.
The solution $t = 0$ represents the time of launch.
For $t-9=0$, we solve for $t$ and get $t=9$.

Answer:

9