6. A diver jumps off a diving board 12 feet above the water with an initial velocity of 20 feet per second and performs two flips. If he wishes to enter the water with as small a splash as possible, he must complete his flips and straighten his body before he is 4 feet above the water. Approximately how much time does the diver have to complete his two flips if he wishes to straighten his body in time to make as small a splash as possible?

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UpStudy AI · Accepted Answer

We model the diver’s vertical motion using the equation

$$
y = y_0 + v_0 t - \frac{1}{2} g t^2
$$

where

- $$y_0 = 12$$ feet (initial height),
- $$v_0 = 20$$ ft/s (initial upward velocity), and
- $$g = 32$$ ft/s$$^2$$ (acceleration due to gravity).

The diver must complete his flips before he reaches a height of 4 feet. Setting $$y = 4$$ yields

$$
4 = 12 + 20t - 16t^2.
$$

Subtract 4 from both sides to simplify the equation:

$$
0 = 12 + 20t - 16t^2 - 4,
$$

$$
0 = 8 + 20t - 16t^2.
$$

Rearrange the terms to obtain a standard quadratic equation:

$$
16t^2 - 20t - 8 = 0.
$$

To simplify, divide the entire equation by 4:

$$
4t^2 - 5t - 2 = 0.
$$

We now use the quadratic formula

$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

with $$a = 4$$, $$b = -5$$, and $$c = -2$$:

$$
t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-2)}}{2 \cdot 4},
$$

$$
t = \frac{5 \pm \sqrt{25 + 32}}{8},
$$

$$
t = \frac{5 \pm \sqrt{57}}{8}.
$$

Since time must be positive, we take the positive solution:

$$
t = \frac{5 + \sqrt{57}}{8}.
$$

Approximating $$\sqrt{57} \approx 7.55$$, we have

$$
t \approx \frac{5 + 7.55}{8} \approx \frac{12.55}{8} \approx 1.57 \text{ seconds}.
$$

Thus, the diver has approximately $$1.57$$ seconds to complete his two flips.
The diver has approximately 1.57 seconds to complete his two flips before reaching 4 feet above the water.

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