Quadratic Formula

Question

A trained stunt diver is diving off a platform that is 15m high into a pool of water that is 45 cm deep. The height, h, in meters, of the stunt diver above the water, is modeled by h= -4.9t^2+1.2t+15, where t is the time in seconds after starting the dive.

a) How long is the stunt diver above 15 m?

b) How long is the stunt diver in the air?

Answer 1

h(t) = -4.9t^2 + 1.2t + 15 = height after t seconds

max height is when h' = 0

h'(t) = -9.8t +1.2 = 0

t = 1.2/9.8 = 12/98 = 6/49 = about 0.12 seconds

for slightly over one tenth of a second, the diver is above 15 feet high

max height = h(6/49) = -4.9(6/49)^2 + 1.2(6/49) + 15 = about 15.0735

from h(0)=15 to h=0 is how long he's above the water

0 = -4.9t^2 + 1.2t + 15

4.9t^2 -1.2t -15 = 0

t= 1.2/9.8 + (1/9.8)sqr(1.2^2 + 4(4.9)1.5)) = about 1.88 seconds