These are so great! They are a perfect combination of Physics and pre-calculus! Your max height of that projectile is going to occur at the max value of the parabola, or at its vertex. So we need to find the vertex. The coordinates of the vertex will give us the x value, which is the time in seconds it takes to reach y which is the max height. Do this by completing the square. Begin by setting the equation equal to 0 and then moving the 80 over to the other side. Then factor out the -16. This is all that: 
. Take half the linear term which is 4 and square it and add it in to both sides. Half of 4 is 2, 2 squared is 4, so add 4 into the set of parenthesis and to the -80. 
. The -64 on the right comes from the fact that when you added 4 into the parenthesis, you had the -16 out in front which is a multiplier. -16 * 4 - -64. So what you really added in was -64. Now the perfect square binomial we created in that process was 
. When we move the 144 back over by addition we find that the vertex of the polynomial is (2, 144). And that tells us that it takes 2 seconds for the projectile to reach its max height of 144 feet. To find the time interval in which the object's height decreases occurs from its max height of 144 to where the graph of the parabola goes through the x-axis to the right of the max. To find where the graph goes through the x-axis, or the zeroes of the graph, you factor the polynomial. When you do that using the quadratic formula you get that x = -1 and 5. So at its max height it is at 2 seconds, and by 5 seconds it hits the ground. So the time interval of its height decreasing is from 2 seconds to 5 seconds, or a total of 3 seconds. I think you need the 2 and 5, from the wording of your problem.
Given that ∠3≅∠5 .
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