Question

Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water in

meters), t seconds after the diver leaves the springboard.

At what time on the divers descent toward the water is the diver again at the same height as the spring board?

Answer 1

2.75 seconds

Explanation:

we are asking nothing else than after how many seconds (after jumping) is the diver at the same height above the water as in his starting position before jumping ?

that is a key factor to understand it that way, because that allows us to calculate the starting point for t=0.

-4t² + 11t + 3 = h(t)

so, for t = 0

h(0) = -4×0² + 11×0 + 3 = 3

so, what do you know, the springboard is actually 3 meters above the water - a 3m springboard.

now, we want to solve this for t under the constriction that the result must be 3. for what values of t is that the case ?

one we know already (t = 0). but there must be a second one (based on the scenario and also on the fact that this is a squared equation).

3 = -4t² + 11t + 3

0 = -4t² + 11t = t×(-4t + 11)

so, we see, t = 0 is one possibility. that was the starting position.

the other one is, when (-4t + 11) = 0.

-4t + 11 = 0

11 = 4t

t = 11/4 = 2.75 seconds.

=>

2.75 seconds after jumping up then falling back down is the diver back at the same height above the water as the diving board.

Answer 2

11/4 seconds

Explanation:

We can see from the given equation that the height at the diving board is h = 3.

This is because the h(t) equation has that +3 at the end, which denotes the initial height of the diver, the height when he is standing on the board before jumping.

To find where h(t) = 3 is true, we need to set h(t) equal to 3 and solve for t.

h(t) = 3

h(t) = -4t^2 + 11t + 3 = 3

-4t^2 + 11t + 3 - 3 = 0

-4t^2 + 11t = 0

t*(-4t + 11) = 0

so h(t) = 3 when t = 0 and when t = 11/4 sec

we already know that at t = 0 the height is 3, it is the initial height given from the equation, so we want to use the other solution for t.

the diver is back at the height of the diving board at t = 11/4 sec