SOLUTION: A model rocket is launched from a platform. The trajectory of the
rocket can be modelled by the relation h=−5t^2+100 t+15, where h is the
height of the model rocket in meters a
Question 1199344: A model rocket is launched from a platform. The trajectory of the
rocket can be modelled by the relation h=−5t^2+100 t+15, where h is the
height of the model rocket in meters and t is the time in seconds.
a) What is the height of the platform?
b) What is the height of the model rocket after 4 seconds?
c) What is the maximum height reached by the rocket? How long does
it take to reach this height?
d) Approximately how long is the rocket in the air? Found 2 solutions by Theo, ikleyn:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! the equation is h = 5^2 + 100t + 15.
h is the height of the rocket.
t is the time in seconds.
the height of the platform is the height when t = 0 which is 15 meters high.
the height of the rocket after 4 seconds is found by replacing t with 4 and solving for h.
you get h = -5 * 4^2 + 100 * 4 + 15 = 335 meters.
the maximum height reached by the rocket is when t = -b/2a.
a = the coefficient of the t^2 term = -5
b is the coefficient of the t term = 100
c is the constant term = 15
t = -b/2a becomes t = -100 / -10 = 10
when t = 10, h = -5 * 10^2 + 100 * 10 + 15 = 515 meters.
it takes 10 seconds for the rocket to reach the maximum height.
the equation can be graphed by letting y represent h and x represent t.
here's what the graph looks like.
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