#### Answer

(a) t = 15.3 seconds
(b) $x = 330~m$
(c) The ball would travel six times farther on the moon.

#### Work Step by Step

(a) We can find the acceleration of gravity on the moon.
$g_m = \frac{9.80~m/s^2}{6} = 1.63~m/s^2$
We can find $t_{up}$, the time for the ball to reach maximum height.
$t_{up} = \frac{v_y-v_{y0}}{g_m}$
$t_{up} = \frac{0-(25~m/s)~sin(30^{\circ})}{-1.63~m/s^2}$
$t_{up} = 7.67~s$
The total time is $2~t_{up}$ which is 15.3 seconds.
(b) We can find the distance $x$ the ball travels on the moon.
$x = \frac{v_0^2~sin(2\theta)}{(g/6)}$
$x = \frac{(6)(25~m/s)^2~sin(60^{\circ})}{9.80~m/s^2}$
$x = 330~m$
(c) We can find the distance the golf ball travels on earth.
$x = \frac{v_0^2~sin(2\theta)}{g}$
$x = \frac{(25~m/s)^2~sin(60^{\circ})}{9.80~m/s^2}$
$x = 55~m$
The ball would travel six times farther on the moon.