Answer to Question #287036 in Physics for omar

Question

A 56-kg diver jumps off the end of a 7.5-m platform with an initial horizontal speed of 

3.6 m/s. 

a) Determine the diver’s total mechanical energy at the end of the platform relative to the surface of the water in the pool below. 

b) Apply the law of conservation of energy to determine the diver’s speed at a height of 2.8 m above the water. 

c) Repeat (b) to find the maximum speed of the diver upon reaching the water.

Answer 1

a) The total mechanical energy is the cum of potential end kinetic energy:

$$E_M=E_P+E_K=mgh+\dfrac12mv^2=4479\text{ J}.$$

b) The potential energy decreases, the kinetic energy increases, keeping the mechanical energy the same:

$$E_{K1}=E_M-E_{P1}=4479-mgy=2942\text{ J},\\\space\\ E_{K1}=\dfrac 12mv_1^2→v_1=\sqrt\dfrac{2E_{K1}}{m}=10.25\text{ m/s}.$$

c) The maximum speed before the surface is

$$v_2=\sqrt\dfrac{2E_{K2}}{m}=\sqrt\dfrac{2(E_M-0)}{m}=12.6\text{ m/s}.$$