Question 167209: Polynomial Word Problem.
A launched rocket has an altitude, in meters, given by the polynomial h+vt-4.92, where h is the height, in meters, from which the launce occurs, at velocity v in meters per second, a t is the number of seconds for which the rocket is airborne. If a rocket is launched from the top of a tower 50 meters high with an initial upward speed of 40 meters per second, what will its height be after 2 seconds?
(round to the nearest tenth)
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A launched rocket has an altitude, in meters, given by the polynomial h+vt-4.92,
where h is the height, in meters, from which the launce occurs, at velocity v
in meters per second, a t is the number of seconds for which the rocket is
airborne. If a rocket is launched from the top of a tower 50 meters high with
an initial upward speed of 40 meters per second, what will its height be after 2 seconds?
(round to the nearest tenth)
:
I think the equation would be: f(t) = -4.92t^2 + vt + h;
where:
f(t) = height in meters after t seconds
-4.92t^2 = force of gravity pulling down
:
f(t) = -4.92t^2 + 40t + 50; equation for this problems
:
To find the height after 2 seconds, substitute 2 for t and and find f(t)
f(t) = -4.92*2^2 40(2) + 50
:
f(t) = -4.92*4 + 80 + 50
:
f(t) = -19.68 + 130
:
f(t) = 110.3 meters after 2 seconds
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