NASA launches a rocket at t =0 t =0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = −4.9 t^2 + 226t + 143 h(t) = -4.9 t^2 + 226t + 143.

Question

NASA launches a rocket at t =0 t =0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = −4.9 t^2 + 226t + 143 h(t) = -4.9 t^2 + 226t + 143. Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? How high above sea-level does the rocket get at its peak?

Answer 1

When rocket splash down when h(t) = - 4.9t²+ 226t + 143 = 0.From here t = 46.75 s;

At the top velocity v(t) = h'(t) = -9.8t + 226 = 0, so t = 226/98 = 23.06 s

Because maximum height is h(23.06) = 2748.92 m.

I used TI-84 for calculations