# College Physics 2.45 – A dolphin in an aquatic show

## Solution:

### Part A

The knowns are

$a=-9.80\:m/s^2$

$v_0=13\:m/s$

$y=0\:m$

### Part B

At the highest point of the jump, the velocity is equal to 0. That is $v=0\:m/s$

Basing from the known values, the formula that we can use to solve for y is $v^2=v_0^2+2a\Delta y$. By rearranging these variables, the formula in solving for $\Delta y$ is $\Delta y=\frac{v^2-v_0^2}{2a}$. Therefore we have,

$\Delta y=\frac{\left(0\:m/s\right)^2-\left(13.0\:m/s\right)^2}{2\left(-9.80\:m/s^2\right)}$

$\Delta y=8.62\:m$

This value is reasonable since dolphins can jump several meters high out of the water. Usually, a dolphin measures about 2 meters and they can jump several times their length.

### Part C

The formula we can use to solve for the time is $v=v_0+at$. If we rearrange this formula and solve for t, it becomes  $t=\frac{v-v_0}{a}$.

$t=\frac{0\:m/s-13.0\:m/s}{-9.8\:m/s^2}$

$t=1.3625\:s$

This value is the time it takes the dolphin to reach the highest point. Since the time it takes to reach this point is equal to the time it takes to go back to water, the time it is in the air is $2.65\:s$.