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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To find the maximum height reached by the ball, we can use the vertex formula for a quadratic function. The height function \( h(t) = -16t^2 + 96t + 5 \) is a downward-opening parabola, and its maximum occurs at \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 96 \). Calculating \( t \): \[ t = -\frac{96}{2(-16)} = \frac{96}{32} = 3 \text{ seconds} \] Now, substitute \( t = 3 \) back into the height function to find the maximum height: \[ h(3) = -16(3)^2 + 96(3) + 5 = -144 + 288 + 5 = 149 \text{ ft} \] So, the maximum height reached by the ball is \( \boxed{149} \text{ ft} \).