Answer
$(a)\space 4.37\space s$
$(b)\space 93.6\space m$
Work Step by Step
(a) Here we use equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the traveled time of the golf ball.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$0=30.3\space m/s\times sin45t^{\circ}+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$0=21.43\space m/s\space t-4.9\space m/s^{2}t^{2}$
$0=t(21.43-4.9\space s^{-1}t)$
Solving this quadratic equation provides two solutions,
$t=0$ or $0=21.43-4.9\space s^{-1}t=>t=4.37\space s$
The first solution represents the situation when the golf ball just begins its flight, so we neglect this one. So, t = 4.37 s
(b) Let's apply equation 3.5a $S=ut+\frac{1}{2}at^{2}$ in the horizontal direction to find the range of the golf ball.
$\rightarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$R=30.3\space m/s\times cos45^{\circ}\space \times 4.37\space s+0$
$R=93.6\space m$
Therefore range of the golfball = 93.6 m