Asked by johnyoun

the path of a rocket fired during a fireworks display is given by the equation s(t)=64t-16t^2 (t to the second power) where t is the time, in seconds, and s is the height, in feet. what is the maximum height, in feet, the rocket will reach? In how many seconds will the rocket hit the ground?

Answers

Answered by bobpursley
This is a parabola. Zeroes (that is s(t) is zero ) when t=0, and t=4.

So the max height due to symettry of the parabola occurs at t=2.
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