In order to find the maximum of the function, we have to find the roots of the first derivative.
[tex]H(t)=-16t^2+vt+s\\ H'(t)=-32t+v=0\\ v=32t\\ t=\frac{v}{32}=\frac{31}{32}=0.97s[/tex]
This gives us the time at which the ball reaches its maximum height.
We don't know s. We can find it because we know that at t=0 the ball is on the ground and its height has to be zero.
[tex]H(0)=0=-16(0)^2+v(0)+s\\ s=0[/tex]
Finally, we can find the maximum height:
[tex]H(0.97)=-16(0.97)^2+31\cdot 0.97=15.0156 \simeq15$ft[/tex]
The answer is C.
