Answer to Question #268918 in Mechanics | Relativity for ara

Question

A 60-kg high-diver starts his dive at a height of 20 m.

a. What is his total mechanical energy at the start of his dive?

b. What is his velocity halfway into the dive?

c. What is his velocity just before hitting the water below?

Answer 1

(a) By the definition of the total mechanical energy, we have:

$$ME_{tot}=KE+PE,$$ $$ME_{tot}=0+PE,$$ $$ME_{tot}=mgh=60\ kg\times9.8\ \dfrac{m}{s^2}\times20\ m=11760\ J.$$

(b) We can find the velocity of the high-diver halfway into the dive from the law of conservation of energy:

$$KE=PE,$$ $$\dfrac{1}{2}mv^2=mgh,$$ $$v=\sqrt{2gh}=\sqrt{2\times9.8\ \dfrac{m}{s^2}\times10\ m}=14\ \dfrac{m}{s}.$$

(c) We can find the velocity of the high-diver just before hitting the water below from the law of conservation of energy:

$$KE=PE,$$ $$\dfrac{1}{2}mv^2=mgh,$$ $$v=\sqrt{2gh}=\sqrt{2\times9.8\ \dfrac{m}{s^2}\times20\ m}=19.8\ \dfrac{m}{s}.$$