Let's compute h'(t) to find its maximum:
[tex]h^{\prime}(t)=-32t+80[/tex]In order to find its maximum, we equal it to 0, solve for t and replace the value we get in h(t):
[tex]-32t+80=0[/tex]From this
[tex]t=\frac{5}{2}[/tex]Substituting in h(t) we get:
[tex]-16(\frac{5}{2}^{})^2+80(\frac{5}{2})+96=196[/tex]Thus the maximum height of the ball will be 196 feet.
In order to know how many seconds will pass until the ball hits the ground, we must solve h(t)=0
[tex]-16t^2+80t+96=0[/tex]From this we get:
[tex]t_1=-1,t_2=6[/tex]Since it doesn't make sense for the ball to hit the ground in negatve time, the ball will hit the ground after 6 seconds.
