## Calculus (3rd Edition)

$$19.6 \mathrm{m}$$
We have the position function \begin{align*} s(t)&=s_{0}+v_{0} t-\frac{1}{2} g t^{2}\\ &=v_{0} t-4.9 t^{2}\end{align*} When the ball hits the ground after $4$ seconds, its height is $0$: \begin{align*} 0&=s(4)\\ &=4 v_{0}-4.9(4)^{2} \end{align*} then $v_{0}=19.6 \mathrm{m} / \mathrm{s} .$ The ball reaches its maximum height when $v=0,$ \begin{align*} v&= v_0-9.8t\\ 0&=19.6-9.8t\\ \end{align*} Hence, $t=2$ At $t=2$, , we have the height \begin{align*} s(2)&=0+19.6(2)-\frac{1}{2}(9.8)(4)\\ &=19.6 \mathrm{m} \end{align*}