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We know two points of the trajectory of the rocket:
1) A height of 5760 ft at time t=8 seconds after launch.
2) A height of 0 ft (landing) at time t=53 seconds after launch.
We also know that the initial position was a height of 0 ft at t=0 seconds.
So we have 3 points to write the equation, that will be a quadratic equation for this kind of trayectory.
As we know we have roots at t=0 and t=53, we can start writing it as:
[tex]h(t)=a(t-0)(t-53)=at(t-53)[/tex]We have one point left, (t,h) = (8, 5760), to find the parameter "a". We can replace t and y in the equation and solve as:
[tex]\begin{gathered} h(t)=at(t-53) \\ 5760=a\cdot8\cdot(8-53) \\ 5760=a\cdot8\cdot(-45) \\ 5760=a\cdot(-360) \\ a=\frac{5760}{-360} \\ a=-16 \end{gathered}[/tex]Then we can write the equation as:
[tex]h=-16t(t-53)=-16t^2+848t[/tex]We can graph it as:
In this kind of trajectories, the maximum height is reached halfway between the launch and the landing.
For any function, we can find the maximum of minimums deriving the function and equal it to 0. We will do it for this function:
[tex]\begin{gathered} \frac{dh}{dt}=-16(2t)+848=0 \\ -32t+848=0 \\ 32t=848 \\ t=\frac{848}{32} \\ t=26.5 \end{gathered}[/tex]The maximum height is reached at time t=26.5 seconds.
Now we can calculate the height at t=26.5 seconds, the maximum height, as:
[tex]\begin{gathered} h(26.5)=-16(26.5)^2+848(26.5) \\ h(26.5)=-16\cdot702.25+22472 \\ h(26.5)=-11236+22472 \\ h(26.5)=11236 \end{gathered}[/tex]Answer:
a) The equation is h(t) = -16t²+848t
b) The maximum height is reached at time t=26.5 seconds.
c) The maximum height is 11236 ft.
