# Falling Objects| College Physics| Openstax| Problem 2.49|

#### You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

SOLUTION:

The known values are

$a=-9.80\:m/s^2;\:v_o=15.0\:m/s;\:y=7.00\:m$

The applicable formula is $y=v_ot+\frac{1}{2}at^2$.

Using this formula, we can solve it in terms of time, t.

$t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}$

Substituting the known values, we have

$t=\frac{-15.0\:m/s\pm \sqrt{\left(15.0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(7.00\:m\right)}}{-9.80\:m/s^2}$

$t=\frac{-15.0\:m/s\pm 9.37\:m/s}{-9.80\:m/s^2}$

We have two values for time, t. These two values represent the times when the ball passes the tree branch. So, the two values of the time, t, are

$t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec$

$t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec$

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds. That is $1.92 seconds$.