Answer:
The rocket splashes down after 24.65 seconds.
The rocket reaches its peak height at 817.74 meters above sea-level.
Step-by-step explanation:
The rocket splashes down after 24.65 seconds.
The rocket reaches its peak height at 817.74 meters above sea-level.
These values were calculated using the given function \( h(t) = -4.9t^2 + 115t + 143 \) by finding the time when the height is zero (splashdown) and the vertex of the parabola (peak height).
Answer:
The rocket splashes down after 24.65 seconds.
The rocket peaks at 817.74 meters above sea-level
Step-by-step explanation:
To find the time when the rocket splashes down, we need to find the value of [tex] t [/tex] when the height ([tex] h(t) [/tex]) equals zero.
We can solve this by setting the equation [tex] h(t) = -4.9t^2 + 115t + 143 [/tex] equal to zero and solving for [tex] t [/tex].
Finding splashdown time ([tex] t [/tex]):
[tex] h(t) = -4.9t^2 + 115t + 143 [/tex]
[tex] 0 = -4.9t^2 + 115t + 143 [/tex]
We can use the quadratic formula to solve for [tex] t [/tex]:
[tex] t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex]
where [tex] a = -4.9 [/tex], [tex] b = 115 [/tex], and [tex] c = 143 [/tex].
[tex] t = \dfrac{-115 \pm \sqrt{115^2 - 4(-4.9)(143)}}{2(-4.9)} [/tex]
[tex] t = \dfrac{-115 \pm \sqrt{13225 + 2808.4}}{-9.8} [/tex]
[tex] t = \dfrac{-115 \pm \sqrt{16033.4}}{-9.8} [/tex]
[tex] t \approx \dfrac{-115 \pm 126.62306267027}{-9.8} [/tex]
We take the positive root since time cannot be negative:
[tex] t \approx \dfrac{-115 + 126.62306267027}{-9.8} [/tex]
[tex] t \approx \dfrac{11.623062670273}{-9.8} [/tex]
[tex] t \approx -1.1860268030891 \textsf{ seconds (ignored since time cannot be negative)} [/tex]
[tex] t \approx \dfrac{-115 - 126.62306267027}{-9.8} [/tex]
[tex] t \approx \dfrac{-241.62306267027}{-9.8} [/tex]
[tex] t \approx 24.655414558191 [/tex]
[tex] t \approx 24.65 \textsf{ seconds (in 2 d.p.)} [/tex]
So, the rocket splashes down after approximately [tex] 24.65 [/tex] seconds.
Finding the peak height:
To find the peak height, we can find the vertex of the parabolic function [tex] h(t) = -4.9t^2 + 115t + 143 [/tex]. The vertex occurs at [tex] t = -\dfrac{b}{2a} [/tex], where [tex] a = -4.9 [/tex] and [tex] b = 115 [/tex].
[tex] t = -\dfrac{115}{2(-4.9)} [/tex]
[tex] t = -\dfrac{115}{-9.8} [/tex]
[tex] t \approx 11.734693877551 [/tex]
[tex] t = 11.73 \textsf{ seconds} [/tex]
Now, substitute [tex] t [/tex] into [tex] h(t) [/tex] to find the peak height:
[tex] h(11.73) = -4.9(11.73)^2 + 115(11.73) + 143 [/tex]
[tex] h(11.73) \approx 817.74479 [/tex]
[tex] h(11.73) \approx 817.74 \textsf{ meters (in 2 d.p.)}[/tex]
So, the rocket peaks at approximately [tex] [/tex] meters above sea level.
