On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The free-fall acceleration on the moon is 1/6 of its value on earth. Suppose he hit the ball with a speed of 25 m/s at an ${30}^{°}$angle above the horizontal.

a. How much farther did the ball travel on the moon than it would have on earth?

b. For how much more time was the ball in flight?

The astronaut hit the golf ball with a 6 iron, the free-fall acceleration on the moon is $\frac{1}{6}$of the its value on earth, the initial speed of the ball is and the launch angle is above the horizontal localid="1650265281344" ${30}^{°}$

The distance travelled by the ball in a medium is localid="1650265284165" $x=\left(u\cos \theta \right)t$

x is the horizontal distance travelled by the ball, u is the initial velocity of the ball, and is the launch angle, and t is the period of flight.

The period of flight is,

localid="1650265286662" $t=\frac{2\left(u\sin \theta \right)}{g}$

The horizontal flight distance is

localid="1650265305282" $x=\left(u\cos \theta \right)\left(\frac{2\left(u\sin \theta \right)}{g}\right)$

For $g=9.8\text{m}/{\text{s}}^2$, the distance travelled by the ball on earth is

$x_2=\left((25\text{m}/\text{s})\cos \left({30}^{°}\right)\right)\left(\frac{2\left((25\text{m}/\text{s})\sin \left({30}^{°}\right)\right)}{9.8\text{m}/{\text{s}}^2}\right)=55.231\text{m}$

For $g=\frac{1}{6}\left(9.8\text{m}/{\text{s}}^2\right)$, the distance travelled by the ball on the moon is

$x_1=\left((25\text{m}/\text{s})\cos \left({30}^{°}\right)\right)\left(\frac{2\left((25\text{m}/\text{s})\sin \left({30}^{°}\right)\right)}{\frac{1}{6}\left(9.8\text{m}/{\text{s}}^2\right)}\right)=331.3872\text{m}$

The distance travelled by the ball on the moon is $\Delta x=x_1-x_2$

$\Delta x=(331.3872\text{m})-(55.231\text{m})=276.15\text{m}$

Therefore, the farther distance travelled by the ball on the moon than it would on earth is $276.15\text{m}$.

The astronaut hit the golf ball with a 6 iron, the free-fall acceleration on the moon is $\frac{1}{6}$of the its value on earth, the initial speed of the ball is $25\text{m}/\text{s}$and the launch angle is ${30}^{°}$above the horizontal.

The period of flight is,

$t=\frac{2\left(u\sin \theta \right)}{g}$

Here, t is the period of flight, g is the acceleration due to gravity, and u is the initial velocity of the ball, and $\theta$ is the angle of launch.

For $g=9.8\text{m}/{\text{s}}^2$, the period of flight of the ball on earth is

$t_2=\frac{2\left((25\text{m}/\text{s})\sin \left({30}^{°}\right)\right)}{9.8\text{m}/{\text{s}}^2}=2.55\text{s}$

Thus, the flight time of the ball on earth is $2.55\text{s}$.

For $g=\frac{1}{6}\left(9.8\text{m}/{\text{s}}^2\right)$, the period of flight of the ball on earth is

$t_1=\frac{2\left((25\text{m}/\text{s})\sin \left({30}^{°}\right)\right)}{\frac{1}{6}\left(9.8\text{m}/{\text{s}}^2\right)}=15.3\text{s}$

The difference in flight time of the ball is,

$\Delta t=t_1-t_2$

$\Delta t=(15.3\text{s})-(2.55\text{s})=12.75\text{s}$

Therefore, the ball will flight $12.75\text{s}$more time in moon than on the earth.