An amateur rocketry club is holding a competition. There is cloud cover at 1050ft. If a rocket is launched with a velocity of 340 ft/s, use the function h(t)=-16t^2+vt+ho to determine how long the rocket is out of sight.

Question

whpalmer4 · Answer

$$h(t)=-16t^2+vt+h_0$$$$h_0 =0$$$$v=340$$

Set h(t) = 1050 and solve to find the values of t where the rocket disappears above the clouds and returns back to visibility, then take the difference to find out how long it was out of sight.

$$1050 = -16t^2+340t+0$$$$t=?$$