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. $E=mgh=60\cdot9.8\cdot20=11760\ (J)$

b. $mgh=mgh/2+mv^2/2\to v=\sqrt{gh}=\sqrt{9.8\cdot20}=14\ (m/s)$

c. $mgh=mv^2/2\to v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot20}=19.8\ (m/s)$