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)=−4.9t2+217t+185. Assuming that the rocket will splash down

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Question 1153455: 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)=−4.9t2+217t+185.
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.

Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (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%28t%29=-4.9t%5E2%2B217t%2B185
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
will occur when h%28t%29=0
0=-4.9t%5E2%2B217t%2B185.........use quadratic formula
t+=+%28-217+%2B-+sqrt%28+217%5E2-4%2A%28-4.9%29%2A185+%29%29%2F%282%2A%28-4.9%29%29+
t+=+%28-217+%2B-+sqrt%28+47089%2B3626+%29%29%2F%28-9.8%29+
t+=+%28-217+%2B-+sqrt%28+50715%29%29%2F%28-9.8%29+
t+=+%28-217+%2B-+225.1999%29%2F%28-9.8%29+
solutions: we need only positive solution
t+=+%28-217+-+225.1999%29%2F%28-9.8%29+
t+=+-%28217%2B225.1999%29%2F%28-9.8%29+
t+=+%28217%2B225.1999%29%2F9.8+
t+=45.1224


The rocket splashes down after ___t+=45.1224_____ seconds.

How high above sea-level does the rocket get at its peak?
h%28t%29=-4.9t%5E2%2B217t%2B185
max occurs when t+=+-b%2F%282a%29=-217%2F%282%28-4.9%29%29=-217%2F-9.8=22.14+
h%2822.14+%29=-4.9%2822.14+%29%5E2%2B217%2A22.14+%2B185
h%2822.14+%29=2587.5
The rocket peaks at ____2587.5______ meters above sea-level.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

In my post, I only want to explain to the visitor, that the given equation DOES NOT describe the flight of a real rocket.


It describe the flight of a PROJECTILE, i.e. an object launched with the initial velocity of 217 m/s,


but then falling freely under the Earth's gravitation force.


The drag force, which provides vertical acceleration of a real rocket,  ABSENTS in the given equation (!)


So, it is not a real rocket (!) (!)


Such object has a common name "a projectile" in Physics.