A ball is thrown striaght up from a height of 3 ft with an initial velocity of 40 feet per second. How high will the ball go? Assume that the only force acting on the ball is gravity where the acceleration due to gravity is -32 feet per second per second.

h(t) = 3 + 40t - 16t^2

Now just find the vertex of the parabola.

To find the height the ball will reach, we can use the kinematic equation:

\[v_f^2 = v_i^2 + 2a \Delta y\]

Where:
- \(v_f\) is the final velocity of the ball, which will be zero when it reaches its highest point,
- \(v_i\) is the initial velocity of the ball,
- \(a\) is the acceleration due to gravity, and
- \(\Delta y\) is the change in height.

In this case, \(v_f = 0\), \(v_i = 40\) ft/s, \(a = -32\) ft/s², and \(\Delta y\) is what we need to find.

Rearranging the equation to solve for \(\Delta y\):

\[2a \Delta y = -v_i^2\]

\[\Delta y = \frac{-v_i^2}{2a}\]

Substituting the given values, we get:

\[\Delta y = \frac{-40^2}{2(-32)}\]

Calculating:

\[\Delta y = \frac{-1600}{-64}\]

\[\Delta y = 25\]

Therefore, the ball will go up to a height of 25 ft.