91Ó°ÊÓ

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" $x=\left(u\cos \mathrm{θ}\right)t$

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" $t=\frac{2\left(u\sin \mathrm{θ}\right)}{g}$

The horizontal flight distance is

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

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 $\mathrm{Δ}x=x_1−x_2$

$\mathrm{Δ}x=(331.3872\text{m})−(55.231\text{m})=276.15\text{m}$

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,

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

Here, t is the period of flight, g is the acceleration due to gravity, and u is the initial velocity of the ball, and $\mathrm{θ}$ 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,

$\mathrm{Δ}t=t_1−t_2$

$\mathrm{Δ}t=(15.3\text{s})−(2.55\text{s})=12.75\text{s}$

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