NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t) = -4.9t^2 + 64 t + 192. Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

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Oddman answered
-4.9t^2 + 64t + 192 = 0 t = (-64 ± √(64^2 - 4(-4.9)(192)))/(2(-4.9)) We are only interested in the positive solution, so we choose t = (64 + √(4096 + 3763.2))/9.8 t ≈ 15.577 The rocket will splash down approximately 15.6 seconds after launch.
Check   -4.9(15.577^2) + 64(15.577) + 192   = -1188.91 + 996.91 + 192   = 0  Note that this assumes the vertical speed at time zero is 64 m/s. If we assume time starts counting at a height of 192 meters after acceleration to that speed, we can compute that acceleration may have taken about 6 seconds. That means the payload will experience an effective gravitational force of about 3.2 times that of Earth's gravity. On the way down, aerodynamic effects will slow descent, so splashdown actually will occur later.

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