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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.


A. when does the diver hit the water?

Answer :

Awhkassgo
The diver will hit the water when the height is zero, so we will want to set our height function equal to zero
0 = -4t^2 + 11t + 3
we will then factor to find out value of t that make the function equal to zero
0 = (-4t - 1 ) ( t - 3 )
This means our function equals zero when t = -1/4 s and t = 3 seconds
Since time cannot be negative, our final solution is t = 3 seconds
Altavistardgo

Answer:

Step-by-step explanation:

When the diver hits the water the height, h(t), becomes zero.  We set h(t) equal to zero and solve the resulting equation for t:

h(t) = -4t^2+ 11t + 3 = 0

The coefficients of this quadratic equation are {-4, 11, 3}, and so the discriminant (needed in the quadratic formula) is -11 - 4(-4)(3) = 47.

                                            -11 ± √37               11 + √37            11 - √37

Therefore the roots are t = --------------, or t = ------------- or t = ---------------

                                                  -8                          8                        8

Both of these results are positive.  Picture the graph as one opening down and intersecting the x-axis in two places.  The diver hits the water (and h becomes zero) at the later time:

       11 + √37

t = ---------------- sec (a little over 2 sec)

              8

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