Hello
The initial height of the ball is [tex]S_i = 15~ft = 4.6~m [/tex], while the initial velocity is [tex]v_i = 40ft/s = 12.2~m/s [/tex].
This is a uniformly accelerated motion, where the acceleration is the gravitational acceleration: [tex]a=-9.8~m/s^2[/tex], where the negative sign means that the acceleration points downwards.
We can proceed in two steps:
1) First of all, we calculate the time the ball needs to reach the maximum height. At the point of maximum height, the velocity is zero: [tex]v_f = 0m/s[/tex].
We can use the formula
[tex]a= \frac{v_f-v_i}{\Delta t} [/tex]
To find the time:
[tex]\Delta t= \frac{v_f-v_i}{a} = 1.2 s[/tex]
2) At this point, we can calculate the total height reached by the ball after this time, i.e. the maximum height:
[tex]S=S_i+v_i \Delta t+ \frac{1}{2} a (\Delta t)^2 =[/tex]
[tex] = 4.6~m+(12.2~m/s)(1.2~s)+ \frac{1}{2}(-9.8~m/s^2)(1.2~s)^2 = 12.1~m[/tex]
