amateur rocketry club is holding a competition. There is cloud cover at 1000 ft. If a rocket is launched with a velocity of 315 ft/s, use the function h(t)=-16t^2+vt+ho to determine how long the rocket is out of sight.

Question

anonymous · Answer

What is the answer?

anonymous · Answer

First set up your equation of motion.  From the data given,$$h(t)=-16t^2+315t$$This assumes we launch from an initial height of zero.  The rocket will be out of sight if it is above the 1000 foot cloud ceiling.  This gives the inequality$$1000  > -16t^2+315t$$

anonymous · Answer

From here we need to solve the inequality.  First step is to compare to zero, so$$\[16t^2-315t+1000<0$$\]Then we solve the related equation to get the critical values.  Consider$$16t^2-315t+1000=0$$Solve that equation, and you have the times where the rocket disappears, and reappears.  The time between is when it is out of sight.