Question

this question is not about solving the stated problem, but about understanding it. a rocket is launched, and its height above sea level t seconds after launch is given by the equation h(t)=-4.9t² + 1400t + 110. a) from what height was the rocket launched? to answer this question, wed find: select an answer b) what is the maximum height the rocket reaches? to answer this question, wed find: select an answer c) if the rocket will splash down in the ocean, when will it splash down? to answer this question, wed find: select an answer question help: message instructor submit question

Explanation:

Step1: Find launch - height

The rocket's height is given by $h(t)=-4.9t^{2}+1400t + 110$. At launch, $t = 0$. Substitute $t = 0$ into the function: $h(0)=-4.9(0)^{2}+1400(0)+110$.

Step2: Calculate the value

$h(0)=110$.

Step3: Find maximum - height

The function $h(t)=-4.9t^{2}+1400t + 110$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-4.9$, $b = 1400$, $c = 110$. The $t$ - value of the vertex of a quadratic function is $t=-\frac{b}{2a}$. So $t=-\frac{1400}{2\times(-4.9)}=\frac{1400}{9.8}$. Then substitute this $t$ - value into $h(t)$ to find the maximum height.

Step4: Find splash - down time

When the rocket splashes down, $h(t)=0$. So we need to solve the quadratic equation $-4.9t^{2}+1400t + 110 = 0$ for $t$. We can use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ where $a=-4.9$, $b = 1400$, $c = 110$.

Answer:

a) The rocket was launched from a height of 110 units.
b) First find $t=\frac{1400}{9.8}$, then substitute into $h(t)$ to get the maximum height.
c) Solve the quadratic equation $-4.9t^{2}+1400t + 110 = 0$ for $t$ using the quadratic formula.