Question

A model rocket is launched with an initial upward velocity of

Answer 1

t = 0.61 or 8.79 The equation you're giving is very badly formatted, but it looks like a variant of h = 47T - 1/2 AT^2 with the 1/2 already factored out and using an A of 10 m/s^2 which is close enough to the actual value of 9.8m/s^2. So I'll use the equation of h = 47t - 5t^2. You're looking for a height of 27 meters. So let's assign that value: h = 47t - 5t^2 27 = 47t - 5t^2 And we can manipulate that to make a quadratic equation. So: 27 = 47t - 5t^2 0 = 47t - 5t^2 - 27 -5t^2 + 47t - 27 = 0 Using the quadratic formula, we can find the roots at 0.61466 and 8.78534 So the rocket is at 27 meters at t = 0.61 seconds as it's going up and at t = 8.79 seconds as it's going back down.... Assuming of course, you don't have a parachute.