# Falling Objects| College Physics| Openstax| Problem 2.45|

#### A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.

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 $2.65\:s$.