# College Physics 2.46 – A swimmer bouncing from a diving board

## 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, $v=-7.16\:m/s$