Question 2.9: MAXIMUM HEIGHT DERIVED GOAL Find the maximum height of a thr...

(a) Find the time it takes the ball to reach its maximum height.

Write the velocity kinematics equation:

v=a t+v_{0}

Move v_{0} to the left side of the equation:

v-v_{0}=a t

Divide both sides by a :

\frac{v – v_{0}}{a}=\frac{\cancel{a} t}{\cancel{a}}=t

Turn the equation around so that t is on the left and substitute v=0, corresponding to the velocity at maximum height:

\text { (1) } t=\frac{-v_{0}}{a}

Replace t by t_{\max } and substitute a=-g :

\text { (2) } t_{\max }=\frac{v_{0}}{g}

(b) Find the maximum height.

Write the equation for the position y at any time:

y= y_{0}+v_{0} t+\frac{1}{2} a t^{2}

Substitute t=v_{0} / a, which corresponds to the time it takes to reach y_{\max }, the maximum height:

y_{\max } =y_{0}+v_{0}\left(\frac{-v_{0}}{a}\right)+\frac{1}{2} a\left(\frac{-v_{0}}{a}\right)^{2}

=y_{0} -\frac{v_{0}^{2}}{a}+\frac{1}{2} \frac{v_{0}^{2}}{a}

Combine the last two terms and substitute a= – g :

(3) y_{\max }=y_{0}+\frac{v_{0}^{2}}{2 g}

REMARKS Notice that g=+9.8 \mathrm{~m} / \mathrm{s}^{2}, so the second term is positive overall. Equations (1) -(3) are much more useful than a numerical answer because the effect of changing one value can be seen immediately. For example, doubling the initial velocity v_{0} quadruples the displacement above the point of release. Notice also that y_{\text {max }} could be obtained more readily from the timeindependent equation, v^{2}-v_{0}^{2}=2 a \Delta y.