Answer
(a) $1.17\ and\ 6.83\sec$
(b) $t=8\ sec$
Work Step by Step
(a)
Step 1. Given $s=-16t^2+v_0t$, with $v_0=128$, we have $s=-16t^2+128t$
Step 2. To reach a height of $80\ ft$, we get $128=-16t^2+128t$ or $t^2-8t+8=0$
Step 3. Let $a=1, b=-8, c=8$, use quadratic formula to solve the above equation and get $t\approx1.17\ and\ 6.83\sec$ (there are two time values for up and down to reach this height.)
(b) For reaching the ground, we have $s=0$, thus $-16t^2+128t=0$ which gives $t=0, 8\ sec$. Discard $t=0$ because it is the starting time, we have $t=8\ sec$
