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:

Move $v_0$ to the left side of the equation:

Divide both sides by a:

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

(1) $t  =  \frac{-v_0}{a}$

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

(2) $t_{max}  =  \frac{v_0}{g}$

(b) Find the maximum height.

Write the equation for the position y at any time:

Substitute $t  =  -v_0/a$, which corresponds to the time it takes to reach $y_{max}$, the maximum height:

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

(3) $y_{max}  =  y_0  +  \frac{v_0²}{2g}$

REMARKS Notice that g = +9.8 m/s², 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 v0 quadruples the displacement above the point of release. Notice also that $y_{max}$ could be obtained more readily from the time-independent equation, $v²  –  v_0² =  2a \Delta y$.