I'm sorry. I don't have time to do all of this, but I can give you a quick rundown.Originally Posted by corey3915
For starters, is both right and wrong. The key to using this equation is that we have a coordinate system, that is, a specified direction to call positive. Most people will define positive to be upward, so we need to modify the equation to reflect that the acceleration due to gravity is downward:
a) Since you are starting by throwing the ball upward, v0 will be positive. Since you are throwing from "ground level" (A common simplification in these problems is that, unless otherwise stated, all objects start from ground height. It's wrong, but that's the way most of these end up reading!), we can put s0 = 0 ft. So your equation will be:
b) Just plug in t = 1 s.
c) The ball will be back at s = 0 ft, so put s = 0 and solve for t.
d) I don't know what level of Math you are at, so let me suggest that either you:
1) Plot the function and graphically figure out where the max height is.
2) Use the fact that the function is a parabola and that the max height will be at the vertex point.
(A third way is to take the first derivative of the s(t) function and set it to zero, but if you don't know Calculus, this suggestion will be gibberish!)