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?


The known values are


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  



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