# A diver springs upward with an initial speed of 1.8 m/sec from a board that is 3.0 m above the water.?

(a) Find the velocity with which he strikes the water.(b) What is the highest point he reaches above the water? Use only formulae in the photo

### 1 Answer

- electron1Lv 79 months ago
In the first equation, Vave = ½ * (vi + vt)

The following equation can be used to calculate the time.

a = (vf – vi) ÷ t

t = (vf – vi) ÷ a

This equation comes from the second equation. The following can be used to calculate the distance an object moves.

d = ½ * (vi + vf) * t

This equation comes from the first equation. Let’s substitute (vf – vi) ÷ a for t.

d = ½ * (vi + vf) * (vf – vi) ÷ a

(vi + vf) * (vf – vi) = 2 * a * d

(vi + vf) * (vf – vi) = vf^2 – vi^2

vf^2 – vi^2 = 2 * a * d

vf^2 = vi^2 + 2 * a * d

This equation can be used to solve both of the problems. a is -9.8 m/s^2. d is the diver’s vertical displacement. d = final height – initial height

For the first problem, the final height is 0 meter. d is -3 meter

vf^2 = 3.24 + 2 * -9.8 * -3

vf = ±√62.04

This is approximately 7.9 m/s. Since the diver is moving downward, its final velocity is negative.

b) What is the highest point he reaches above the water? Use only formulae in the photo?

At the highest point, its vertical velocity is 0 m/s. Let’s use the same equation.

0 = 3.24 + 2 * -9.8 * d

19.6 * d = 3.24

d = 3.24 ÷ 19.6

This is approximately 0.17 meter. These are the same answers. But, I think this is an easier way to solve these type of problem.

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