#### Answer

(a) The flight time is 15.3 seconds.
(b) $x = 331~m$
(c) The ball would travel 276 meters 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.
$t_{up} = \frac{v_{y0}}{a}$
$t_{up} = \frac{(25~m/s)~sin(30^{\circ})}{1.63~m/s^2}$
$t_{up} = 7.67~s$
The total time is $2~t_{up}$ which is 15.3 seconds.
(b) $x = v_x~t$
$x = (25~m/s)~cos(30^{\circ})(15.3~s)$
$x = 331~m$
(c) We can find the range of the golf ball on earth.
$range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{(25~m/s)^2~sin(60^{\circ})}{9.80~m/s^2}$
$range = 55~m$
The ball would travel 276 meters farther on the moon.