Asked by Connor
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.
When does the diver hit the water, and what was his velocity?
When does the diver hit the water, and what was his velocity?
Answers
Answered by
Damon
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
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