question 8\nscore on last try: 1 of 2 pts. see details for more.\nat least one scored part is incorrect…
question 8\nscore on last try: 1 of 2 pts. see details for more.\nat least one scored part is incorrect. jump to first changable incorrect part.\nget a similar question you can retry this question below\nnasa launches a rocket at t = 0 seconds. its height, in meters above sea - level, as a function of time is given by h(t)=-4.9t² + 82t + 409.\nassuming that the rocket will splash down into the ocean, at what time does splashdown occur?\nthe rocket splashes down after 20.76 seconds.\nhow high above sea - level does the rocket get at its peak?\nthe rocket peaks at 762.05 meters above sea - level.\nquestion help: video 1 video 2 message instructor\nsubmit question
Answer
Explanation:
Step1: Identify the function type
The height - function $h(t)=-4.9t^{2}+82t + 409$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-4.9$, $b = 82$, and $c = 409$.
Step2: Find the time of the peak
The $t$ - value of the vertex of a quadratic function $y = ax^{2}+bx + c$ is given by $t=-\frac{b}{2a}$. Substitute $a=-4.9$ and $b = 82$ into the formula: $t=-\frac{82}{2\times(-4.9)}=\frac{82}{9.8}\approx8.37$ seconds.
Step3: Find the height at the peak
Substitute $t = \frac{82}{9.8}$ into the height - function $h(t)=-4.9t^{2}+82t + 409$. [ \begin{align*} h\left(\frac{82}{9.8}\right)&=-4.9\times\left(\frac{82}{9.8}\right)^{2}+82\times\frac{82}{9.8}+409\ &=-4.9\times\frac{6724}{96.04}+\frac{6724}{9.8}+409\ &=-\frac{32947.6}{96.04}+\frac{6724}{9.8}+409\ &=- 343+686+409\ &=752 \end{align*} ]
Answer:
The rocket peaks at 752 meters above sea - level.