possible A ball is thrown directly upward from a height of 8 ft with an initial velocity of \( 20 \mathrm{ft} / \mathrm{sec} \). The function \( s(t)=-16 \mathrm{t}^{2}+20 \mathrm{t}+8 \) gives the height of the ball, in feet, \( t \) seconds after it has been thrown. Determine the time at which the ball reaches its maximum height and find the maximum height.

To find the time at which the ball reaches its maximum height, we can use the vertex formula for a quadratic equation. The time at which the maximum height occurs is given by \( t = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the equation \( s(t) = -16t^2 + 20t + 8 \). Here, \( a = -16 \) and \( b = 20 \). Calculating this gives: \[ t = -\frac{20}{2 \times -16} = \frac{20}{32} = \frac{5}{8} \, \text{seconds} \] To find the maximum height, substitute \( t = \frac{5}{8} \) back into the height equation: \[ s\left(\frac{5}{8}\right) = -16\left(\frac{5}{8}\right)^2 + 20\left(\frac{5}{8}\right) + 8 \] \[ = -16 \times \frac{25}{64} + 20 \times \frac{5}{8} + 8 \] \[ = -\frac{400}{64} + \frac{100}{8} + 8 \] \[ = -\frac{400}{64} + \frac{800}{64} + \frac{512}{64} \] \[ = \frac{912 - 400}{64} = \frac{512}{64} = 8 \] So, the maximum height is \( 8 \, \text{ft} \) and it occurs at \( t = \frac{5}{8} \, \text{seconds} \).