Andrew dives off a jetty. His height above the water is t seconds after diving m is given by H(t) = −4t 2 + 4t + 3 metres, t ≥ 0.

Question

2 + 4t + 3 metres, t ≥ 0.

a How high above the water is the jetty?

b How long does it take for Andrew to reach the maximum height of hid dive?

c How far is Andrew above the water at his highest point?

d How long does it take for Andrew to hit the water?

Answer 1

H(t) = -4t^2 +4t +3

take the derivative and set equal to zero

H'(t) = -8t +4 =0

8t = 4

t = 4/8 = 1/2 second to reach maximum height

H(1/2) = -4(1/2)^2 + 4(1/2) +3 =-4(1/4) + 2 + 3 = 4 feet = max height

he hits the water when H(t) = 0

-4t^2 + 4t + 3 = 0,

4t^2 -4t -3 = 0

(2t -3)(2t+1) = 0

2t-3 = 0

t= 3/2 = 1.5 seconds to hit the water