SOLUTION: NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)= -4.9t^2 + 196t + 311 Assuming that the rocket will splash

Question 1200505: NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)= -4.9t^2 + 196t + 311
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after [ ] seconds.
How high above sea-level does the rocket get at its peak?
The rocket peaks at [ ] meters above sea-level.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)= -4.9t^2 + 196t + 311
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after [ ] seconds.
How high above sea-level does the rocket get at its peak?
The rocket peaks at [ ] meters above sea-level.
--------------------
1st, that's not a rocket, it's a projectile.
h(t)= -4.9t^2 + 196t + 311
The max of the parabola is at t - -b/2a = -196/(-4.9*2) = 20 seconds
h(20) = -4.9*400 + 196*20 + 311 = 2271 meters MSL
===========================
The rocket splashes down after? seconds.
s = at^2/2
2271 = 4.9t^2
t =~ 21.53 seconds

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