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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