Question 174670
You can use the function for the height of an object propelled upwards with an initial velocity of {{{v[0]}}} from an initial height of {{{h[0]}}}:
{{{h(t) = -16t^2+v[0]t+h[0]}}}
In this problem, {{{h[0] = 4}}} and {{{v[0] = 25}}}, so, making the substitutions, we get:
{{{h(t) = -16t^2+25t+4}}}
You can find the t-value of the vertex by:
{{{t = -b/2a}}} where: b = 25 and a = -16, so:
{{{t = (-25)/2(-16)}}}
{{{t = 25/32}}} 
Let's look at the graph of this function:
{{{graph(400,400,-5,5,-5,15,-16x^2+25x+4)}}}
The vertex is the point at the top of the curve.
Since we already have the t-value of the vertex, we can find the h-value by substituting {{{t = 25/32}}} into the above equation and we get {{{h = 881/64}}}
So the vertex is located at (0.78, 13.77)