# A ball is thrown directly upward from height 6 ft with initial velocity of 20 ft/sec. The function s(t) = -16t^2 + 20t + 6 gives height of the ball t seconds after thrown. Determine the time where the ball reaches its max height and find the max height?

Oct 24, 2017

The ball reaches its maximum height of $12 \frac{1}{4}$ feet after $\frac{5}{8}$ seconds.

#### Explanation:

As at $t = 0$, $s \left(t\right) = 6$ and as $v \left(t\right) = s ' \left(t\right) = - 32 t + 20$, the initial velocity $v \left(0\right) = 20$. Hence these conditions are built within the function. As such we can proceed with the function to find maximum height.

The function $s \left(t\right) = - 16 {t}^{2} + 20 t + 6$ gives its maxima when converted to vertex form. So doing so

$s \left(t\right) = - 16 {t}^{2} + 20 t + 6$

= $- 16 \left({t}^{2} + \frac{20}{16} t\right) + 6$

= $- 16 \left({t}^{2} + \frac{5}{4} t\right) + 6$

= $- 16 \left({t}^{2} + 2 \times \left(\frac{5}{8}\right) t + {\left(\frac{5}{8}\right)}^{2}\right) + 16 {\left(\frac{5}{8}\right)}^{2} + 6$

= $- 16 {\left(t - \frac{5}{8}\right)}^{2} + \frac{25}{4} + 6$

= $- 16 {\left(t - \frac{5}{8}\right)}^{2} + 6 \frac{1}{4} + 6$

= $- 16 {\left(t - \frac{5}{8}\right)}^{2} + 12 \frac{1}{4}$

Hence, the ball reaches its maximum height of $12 \frac{1}{4}$ feet after $\frac{5}{8}$ seconds.