If a football is kicked straight up with an initial velocity of $$ 96 \mathrm{ft} / \mathrm{sec} $$ from a height of 5 ft , then its height above the earth is a function of time given by $$ \mathrm{h}(\mathrm{t})=-16 \mathrm{t}^{2}+96 \mathrm{t}+5 $$. What is the maximum height reached by the ball? $$ \square \mathrm{ft} $$

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We are given the function for height:

$$
h(t) = -16t^2 + 96t + 5
$$

**Step 1: Find the time of maximum height**

Since the expression is a quadratic with a negative coefficient for $$t^2$$, the function reaches a maximum at its vertex. The time $$t$$ at the vertex is determined by:

$$
t = -\frac{b}{2a}
$$

with $$a = -16$$ and $$b = 96$$. Substituting we get:

$$
t = -\frac{96}{2(-16)} = -\frac{96}{-32} = 3 \text{ seconds}
$$

**Step 2: Compute the maximum height**

Substitute $$t = 3$$ back into the height function:

$$
h(3) = -16(3)^2 + 96(3) + 5
$$

Calculate each term:

$$
-16(9) = -144, \quad 96(3) = 288
$$

Then,

$$
h(3) = -144 + 288 + 5 = 149 \text{ ft}
$$

Thus, the maximum height reached by the ball is $$ \boxed{149 \text{ ft}} $$.
The maximum height reached by the ball is 149 feet.

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