Answer to Question #298329 in Mechanics | Relativity for MishaTA

Question

A diver of mass 60 Kg dives from a height of 10 m into a swimming pool. Through the air there is resistance of 50 N a) Find the speed at which the diver enter the water b) Once in the water, the water provides an upward force of 2000 N. Find the greatest depth in water the diver reaches.

Answer 1

Explanations & Calculations

• Assuming only the given friction forces to apply throughout the motion.

a)

• Consider energy conversion with conservation, the potential to friction work +kinetic energy.

$\qquad\qquad \begin{aligned} \small mg&H=\small fH+\frac{1}{2}mv^2\\ \small v&=\small \sqrt{\frac{2(mg-f)H}{m}} \end{aligned}$

b)

• For this also, energy conservation can be regarded; initial kinetic energy at the waterline into work against upthrust ($\small f_2.h$ ) + loss in potential energy with respect to the waterline.

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mv^2&=\small f_2.h-mgh \end{aligned}$

• Substituting the kinetic energy from the first relation & is possible as kinetic energy gain is just an intermediate phenomenon.

$\qquad\qquad \begin{aligned} \small mgH-f_1H&=\small f_2h-mgh\\ \small h&=\small \frac{(mg-f)H}{(f_2-mg)} \end{aligned}$