The rocket splashes down after 43 seconds.
The rocket peaks at 2400. meters above sea-level.
Step 1
We realise that the rocket will hit the ground when our function [tex]h(t)=0[/tex] , i.e [tex]-4.9t^2+202t+318=0.[/tex]
Step 2
Use the quadratic formula to solve for the time. For the general quadratic equation [tex]at^2+bt+c=0[/tex], the general solution to the equation is
[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} .[/tex]
In our case, the quadratic equation [tex]h(t)=-4.9t^2+202t+318=0[/tex] has a solution
[tex]t=\frac{-202\pm\sqrt{202^2-4((-4.9)(318))}}{2(-4.9)} \\t=-1.52,t=42.74.[/tex]
Step 3
In this step we pick the value of [tex]t[/tex] that makes physical sense. Since the NASA team launced the rocket at [tex]t=0[/tex], the rocket reaches the bottom when [tex]t=42.74[/tex]. This can be rounded up to [tex]43[/tex] seconds.
Step 1
The first step is to realist that since the coefficient of the quadratic term is negative, the graph of this function will face downwards. From this we can gather that the maximum value of this function occurs at the vertex.
Step 2
In this step we calculate the t coordinate of the of the vertex of this function. For any quadratic equation [tex]h(t)=at^2+bt+c[/tex] , the vertex occurs when ,
[tex]t=-\frac{b}{2a}.[/tex]
For the equation [tex]h(t)=-4.9t^2+202t+318[/tex] , the vertex occurs when ,
[tex]t=-\frac{202}{2(-4.9)}=20.61.[/tex]
Step 3
We substitute the value of [tex]t[/tex] from step 2 above into [tex]h(t)[/tex] to get the maximum value of this function. This calculation is shown below,
[tex]h(20.61)=-4.9(20.61)^2+202(20.61)+318=2399.84m.[/tex]
This can be rounded up to 2400m.
