Question

A diver springs upward with an initial speed of 2.90 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

Answer 1

(a) The height the diver will reach is

$$h=\frac{v^2}{2g}.$$

The speed at the impact is

$$u=\sqrt{2gH}=\sqrt{2g(h+3)}=8.2\text{ m/s}.$$

(b) The highest point is H:

$$H=h+3=h=\frac{v^2}{2g}+3=3.42\text{ m}.$$