Joshua jumps from a platform diving board that is 32 feet above the water. The position of the diver can be modeled by s(t) = -16t2+16t+32, where s is his position and t is the seconds that have gone by since he jumped.

Question

When does the diver hit the water, and what was his velocity?

Answer 1

Oh good grief, feet :( old text

had initial speed 16 ft/s up ? I guess ?

h = 32 + 16 t - 16 t^2
0 = 32 + 16 t - 16 t^2
solve quadratic

t = [ -16 +/- sqrt ( 256+2048) ] /-32
t = 2 seconds

v = dh/dt = 16 - 32 t
= 16 - 64
= -48 ft/s