Answer
$0.26$ m
Work Step by Step
A particle's horizontal range $R$,
which is the horizontal distance from the launch point to the point at which the particle returns to the launch height, is
$R=\displaystyle \frac{v_{0}^{2}}{g}\sin 2\theta_{0}$. $\qquad$ (4-26)
The range is greatest when $\theta_{0}=45^{o}$ , in which case $\sin 2\theta_{0}=1$
$R=\displaystyle \frac{v_{0}^{2}}{g}=\frac{(9.50\mathrm{m}/\mathrm{s})^{2}}{9.80\mathrm{m}/\mathrm{s}^{2}}=9.21$ m
This is only 0.26 m = 26 cm more than Powell's result.
