Posted by Anonymous on Wednesday, May 30, 2012 at 12:30am.
paul launches a rocket from a base standing 2 feet off the ground. the rocket has an initial velocity of 50 feet per second. how long will it take for the rocket to reach 100 ft. high?

geometry  Pendergast, Wednesday, May 30, 2012 at 12:36am
I assume the the only forces on the rocket after launch are due to gravity. Gravity exerts an acceleration of 32ft/s^2.
The height function h(t), derived from acceleration a(t) = 32ft/s^2, is h(t) = (1/2)(32ft/s^2)*t^2 + v_0*t + h_0. v_0 is the initial velocity, and h_0 is the initial height. Since this requires calculus to derive, I assume your teacher provided it.
Thus, the equation for our rocket is h(t) = (16ft/s^2)*t^2 + (50ft/s)*t + 2ft. Find the intersection with h = 100ft.
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