Given the function:
[tex]h(t)=-4.9t^2+136t+311[/tex]Given that NASA launches a rocket at t= 0 seconds.
Where h is the height in meters above sea level.
Let's solve for the following:
• (a). The time the rocket splashes the water.
When the rocket splashes the water, the height above sea-level will be 0 meters.
To solve for t, set h(t) for 0 and solve.
We have:
[tex]-4.9t^2+136t+311=0[/tex]Now, let's solve using the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Where:
a = -4.9
b = 136
c = 311
Plug in the values into the formula and solve for t:
[tex]\begin{gathered} t=\frac{-136\pm\sqrt{136^2-4(-4.9)(311)}}{2(-4.9)} \\ \\ t=\frac{-136\pm\sqrt{18496-(-6095.6)}}{-9.8} \\ \\ t=\frac{-136\pm\sqrt{18496+6095.6}}{-9.8} \\ \\ t=\frac{-136\pm\sqrt{24591.6}}{-9.8} \\ \\ t=\frac{-136\pm156.82}{-9.8} \\ \\ \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} t=\frac{-136-156.82}{-9.8},\frac{-136+156.82}{-9.8} \\ \\ t=\frac{−292.82}{-9.8},\frac{20.82}{-9.8} \\ \\ t=29.88,-2.12 \end{gathered}[/tex]Thus, we have the solutions:
t = 29.88 and t = -2.12
Since the time cannot be negative, let's take the positive solution.
Therefore, the rocket splashes down after 29.88 seconds.
• (b). Let's find the peak.
To find the maximum time, apply the fomula:
[tex]t=-\frac{b}{2a}[/tex]Thus, we have:
[tex]\begin{gathered} t=-\frac{136}{2(-4.9)} \\ \\ t=-\frac{136}{-9.8} \\ \\ t=\frac{136}{9.8} \\ \\ t=13.88 \end{gathered}[/tex]The rocket gets to its peak at 13.88 seconds.
Now, to find the height at that time, substitute 13.88 for t in h(t) and solve for h(13.88):
[tex]\begin{gathered} h(13.88)=-4.9(13.88)^2+136(13.88)+311 \\ \\ h(13.88)=-944.01+1887.35+311 \\ \\ h(13.88)=1254.67\text{ m} \end{gathered}[/tex]Therefore, the rocket peaks at 1254.67 above seal level.
ANSWER:
• 29.88 seconds
• 1254.67 meters
