an object is launched from ground level directly upward at a rate of 144 feet per second. the equation for…

an object is launched from ground level directly upward at a rate of 144 feet per second. the equation for the objects height, h, is (h(t)=-16t^{2}+144t), where t is the time in seconds. how long was the object at a height of at least 288 feet?\na. 6 seconds\nb. 9 seconds\nc. 3 seconds\nd. 4.5 seconds
Answer
Explanation:
Step1: Set up the inequality
We want to find when $h(t)\geq288$, so $- 16t^{2}+144t\geq288$. Rearrange it to $16t^{2}-144t + 288\leq0$. Divide through by 16 to get $t^{2}-9t + 18\leq0$.
Step2: Factor the quadratic
Factor $t^{2}-9t + 18$ as $(t - 3)(t - 6)\leq0$.
Step3: Find the roots
Set $(t - 3)(t - 6)=0$. The roots are $t = 3$ and $t = 6$.
Step4: Determine the solution interval
The quadratic $y=(t - 3)(t - 6)$ is a parabola opening upwards. The inequality $(t - 3)(t - 6)\leq0$ is satisfied when $3\leq t\leq6$.
Step5: Calculate the time - duration
The time the object is at a height of at least 288 feet is $6 - 3=3$ seconds.
Answer:
C. 3 seconds