Question

Suppose h(t)=−5t2+11t+13

h
(
t
)
=
-
5
t
2
+
11
t
+
13
represents the height of a diver above the water, in meters, t seconds after the diver leaves the springboard. Approximately when does the diver hit the water? Round your answer to the nearest tenths place.

Answer 1

3.1 seconds

Explanation:

The diver will hit the water when the height is zero.

0 = -5t^2 +11t +13

Using the quadratic formula, we find the solutions to be ...

t = (-11 ±√(11^2 -4(-5)(13)))/(2(-5)) = (-11 ±√381)/-10

The positive solution is

t = 1.1 +√3.81 ≈ 3.1 . . . seconds

The diver will hit the water after 3.1 seconds.