Atoy rocket is launched from the top of a building 117 feet tall at an initial velocity of 201 feet per second.
a) give the function that describes the height of the rocket in terms of time t.
b) determine the time at which the rocket reaches its maximum height, and the maximum height in feet.
c) for what time interval will the rocket be more than 499 feet above ground level
(d) after how many seconds will it hit the ground?

Explanation:

The distance equation is normally d = v0*t + .5at^2 with a of course being the acceleration due to gravity, so -32 ft/s^2.  The problem gives us v0, but also tells us we are already starting 117 feet, so we want to add a 117 to that.  so d = v0*t + .5at^2 + 117.  And this is in terms of t so that's what is asked for.

If you need extra help finding the maximum let me know, but completeing the square can get you the vertex form which makes finding the maximum very easy.

If you set it equal to 499 and solve for t it will get you the two times that it is at 499 feet, and that will get you the time interval.

If you simply set it to 0 it will get you the two times when the function equals 0, one should be positive and the other negative, since you can't go into negative time the positive one will be when it hits the ground.

Again, let me know if you need help carrying out these calculations and I'll be happy to walk you through them.  

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.