## Algebra 2 (1st Edition)

$\sqrt{2.5}$s
The height function is $h=-16t^2+h_0$ where $h_0$ is the initial height and $t$ is the passed time. Hence here: $h=-16t^2+40$. He lands when $h=0$, thus: $0=-16t^2+40\\16t^2=40\\t^2=2.5\\t=\pm\sqrt{2.5}$ But $t$ must be positive, thus $t=\sqrt{2.5}$, and this is how long he was in the air for.