Apollo astronauts took a "nine iron" to the Moon and hit a golf ball about 180 m.
Part A
Assuming that the swing, launch angle, and so on, were the same as on Earth where the same astronaut could hit it only 32 m, estimate the acceleration due to gravity on the surface of the Moon. (We neglect air resistance in both cases, but on the Moon there is none.)
Express your answer to two significant figures and include the appropriate units.
gMoon =

Respuesta :

The acceleration due to gravity on the surface of the Moon is 1.7 m/s²

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

Given:

[tex]d_{moon} = 180 ~ m[/tex]

[tex]d_{earth} = 32 ~ m[/tex]

[tex]g_{earth} = 9.8 ~ m/s^2[/tex]

Unknown:

[tex]g_{moon} =[/tex] ?

Solution:

The motion of the golf ball is a parabolic motion, then we can use the following formula to calculate the horizontal displacement of the ball.

[tex]\large {\boxed {d = \frac{u^2 \sin 2 \theta}{g}} }[/tex]

Let's use this formula to find a comparison of the motion of golf ball on earth and on the moon.

[tex]d_{moon} : d_{earth} = \frac{u^2 \sin 2 \theta}{g_{moon}}} : \frac{u^2 \sin 2 \theta}{g_{earth}}}[/tex]

[tex]d_{moon} : d_{earth} = \frac{1}{g_{moon}}} : \frac{1}{g_{earth}}}[/tex]

[tex]d_{moon} : d_{earth} = g_{earth} : g_{moon}[/tex]

[tex]180 : 32 = 9.8 : g_{moon}[/tex]

[tex]g_{moon} = (32 \times 9.8) \div 180[/tex]

[tex]\large {\boxed {g_{moon} = 1.7 ~ m/s^2} }[/tex]

Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Sperm , Whale , Travel

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