College Physics by Openstax Chapter 2 Problem 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/s2a=-9.80\:\text{m/s}^2; vo=15.0 m/sv_o=15.0\:\text{m/s}; y=7.00 my=7.00\:\text{m}

The applicable formula is.

y=vot+12at2y=v_ot+\frac{1}{2}at^2

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

t=−v0±v02+2ayat=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

t=−v0±v02+2ayat=−15.0 m/s±(15.0 m/s)2+2(−9.80 m/s2)(7.00 m)−9.80 m/s2t=−15.0 m/s±9.37 m/s−9.80 m/s2\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\ t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\ t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2} \end{align*}

We have two values for time, tt. These two values represent the times when the ball passes the tree branch.

t1=−15.0 m/s+9.37 m/s−9.80 m/s2=0.57 sect2=−15.0 m/s−9.37 m/s−9.80 m/s2=2.49 sec 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.

t2−t1=2.49 s−0.57 s=1.92 s  (Answer)t_2-t_1=2.49 \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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