Question 174670
For motion in the vertical direction, the acceleration due
to gravity has to be included in the equation of motion
Motion in the horizontal direction is not included in this
problem.
The equation for motion in the vertical direction (on planet earth)is
{{{y = v[0]t - (1/2)*32t^2}}}
Where {{{v[0]}}} is the initial velocity of the baton, so
{{{y = 25t - 16t^2}}}
There is an initial height also, of 4 ft, so the equation becomes
{{{y = 25t - 16t^2 + 4}}}
Here's a plot. {{{y}}} is the vertical axis and {{{t}}} is the horizontal axis
{{{ graph( 500, 500, -1, 3, -1, 16, -16x^2 + 25x + 4) }}}
The graph shows the situation correctly
At {{{t=0}}},{{{y=4}}} as it should
The baton takes the same amount of time to go from {{{y=4}}} to it's
peak as it does to go from the peak back to {{{y=4}}}. Then it continues
on to the ground.
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The vertex is midway between the roots of the equation. You can
see by the graph that one root is negative and one is positive. the
vertex ends up being at {{{t = (-b)/(2a)}}} when the equation is in the
form {{{y = at^2 + bt + c}}}. In our equation 
{{{a = -16}}}
{{{b = 25}}}
{{{c = 4}}}
{{{(-b)/(2a) = (-25)/(-32)}}}
{{{(-25)/(-32) = 25/32}}}
This is the time in seconds that it takes to reach the vertex
Now plug this back into the equation
{{{y = 25t - 16t^2 + 4}}}
{{{y = 25*(25/32) - 16*(625/1024) + 4}}}
{{{y = 19.53125 - 9.765625 + 4}}}
{{{y = 13.765625}}}ft
This is verified by the graph, but you didn't really
need this- it's extra info.