# Falling Objects| College Physics| Openstax| Problem 2.46|

#### A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water?

SOLUTION:

The known values are: $y_0=1.80\:m,\:y=0\:m,\:a=-9.80\:m/s^2,\:v_0=4.00\:m/s$

Part a

Based from the knowns, the formula most applicable to solve for the time is  $\Delta y=v_0t+\frac{1}{2}at^2$. If we rearrange the formula by solving for t, and substitute the given values, we have

$t=\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a}$

$t=\frac{-4.00\:m/s\pm \sqrt{\left(4.00\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.80\:m\right)}}{-9.80\:m/s^2}$

$t=1.14\:s$

Part b

We have the formula

$\Delta y=\frac{v^2-v_0^2}{2a}$

$\Delta y=\frac{\left(0\:m/s\right)^2-\left(4.00\:m/s\right)^2}{2\left(-9.80\:m/s^2\right)}$

$\Delta y=0.816\:m$

Part c

The formula to be used is

$v^2=v_0^2+2a\Delta y$

$v=\pm \sqrt{v_0^2+2a\Delta y}$

$v=\pm \sqrt{\left(4.00\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-1.80\:m\right)}$

$v=\pm \sqrt{51.28\:m^2/s^2}$

$v=\pm 7.16\:m/s$

Since the diver must be moving in the negative direction, $v=-7.16\:m/s$