# Quadratics

The path of a bottle rocket being launched into the air and falling back to the ground can be modeled by the equation s=1/2 at^2+v_o t+s_o where s is the height of the object, a is the acceleration, vo is the initial velocity, and so is the initial height. For one launch, the following measurements were taken at t = 0.5 second the height was measured as 144 feet, at t = 1 second the height was 152 feet, and at t = 2.5 seconds the height was 128 feet. Use this information to answer the following questions.
What was the initial velocity of the rocket?
What was the initial height of the rocket?
What was the acceleration?
What was the maximum height the rocket reached?
When did the rocket hit the ground?

I am at a loss on how to get started.

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asked by Lost
1. you have three equations and three unknowns
... the unknowns are ... a, Vo, and So

144 = 1/2 a (.5^2) + Vo .5 + So

152 = 1/2 a (1^2) + Vo 1 + So

128 = 1/2 a (2.5^2) + Vo 2.5 + So

solve the system for a, Vo, and So
... use the resulting equation to find max height and flight time
... max height occurs at max time (on the axis of symmetry ... -b/2a)
... flight time is when s equals zero

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posted by R_scott

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