a. The rocket splashes down after 63.94 seconds.
b. The rocket peaks at 7386.68 m
a. How to find when the rocket splashes down?
Since the height of the rocket is h(t) = -4.9t² + 370t + 402, the rocket splashes down when h(t) = 0.
So, h(t) = 0
-4.9t² + 370t + 402 = 0
Using the quadratic formula, we find t.
[tex]t = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}[/tex]
where a = -4.9, b = 370 and c = 402
So, [tex]t = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}\\= \frac{-307 +/- \sqrt{(307)^{2} - 4(-4.9)(402)} }{2(-4.9)}\\= \frac{-307 +/- \sqrt{94249 + 7879.2} }{-9.8)}\\= \frac{-307 +/- \sqrt{102128.2} }{-9.8)}\\= \frac{-307 +/- 319.58}{-9.8)}\\= \frac{-307 + 319.58}{-9.8)} or \frac{-307 - 319.58}{-9.8)}\\= \frac{12.58}{-9.8)} or \frac{-626.58}{-9.8}\\= -1.28 or 63.94[/tex]
Since t cannot be negative t = 63.94 s
So, the rocket splashes down after 63.94 seconds.
b. How to find the peak of the rocket?
Since h(t) = -4.9t² + 370t + 402, to find the time the rocket reaches it peak, we differentiate h(t) with respect to t and equate to zero.
Soi, dh(t)/dt = d(-4.9t² + 370t + 402)/dt
= -9.8t + 370
dh(t)/dt = 0
-9.8t + 370 = 0
-9.8t = -370
t = -370/-9.8
t = 37.76 s
Substitung t = 37.76 into h(t), we have
h(t) = -4.9t² + 370t + 402
h(37.76) = -4.9(37.76)² + 370(37.76) + 402
h(37.76) = -4.9(1425.82) + 370(37.76) + 402
h(37.76) = -6986.518 + 13971.2 + 402
h(37.76) = 7386.682 m
h(37.76) ≅ 7386.68 m
So, the rocket peaks at 7386.68 m
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