How long will the projectile stay above a given height?

Given the initial velocity of a projectile launched on a level surface, we calculate how long the projectile will stay above a given height.
The conceptual key to this problem is to realize that only a vertical analysis of the projectile motion is necessary because we don't care how far the projectile moves horizontally, only how long it takes to pass through the given height at two different times.
So we start by finding the initial y velocity of the projectile, then we write down the y coordinate as a function of time (the vertical position equation for the projectile): y=y_0+v_0yt-1/2gt^2. We sub in the given height for y, and we are left with a quadratic equation in t. This is a good thing, because we'll get two solutions for time: those are the two times for which the projectile passes through the given height!
We solve for the times given the height, and we find the difference of those times in order to get the total time the projectile stays above the given height.

[@shfunko]

Neat stuff!

[@ZaksLab]

thanks! z