Question 389336
The man jumped at 480 feet (t=0, also if you know elementary physics you know that with constant acceleration, the position X is given as {{{(1/2)at^2 + (v_0)t + x_0}}} where x_0 is the original position)


The highest point he reached occurred at the vertex, which is at t = -b/2a = 1/2 second. Plugging in t = 1/2,


h(1/2) = -16(1/4) + 16(1/2) + 480 = 484 feet


To find the amount of time it took to hit the ground, we plug h(t) = 0 and solve:


{{{0 = -16t^2 + 16t + 480}}}


I'll divide both sides by -16


{{{0 = t^2 - t - 30}}}


{{{0 = (t - 6)(t + 5)}}}


{{{t = 6}}} (seconds)