# Solve the Physics Problem: Astronaut Drops Camera on Moon

• lbutscha
In summary, during a walk on the Moon, an astronaut drops his camera over a 14.7 m cliff and it attains a velocity of 3.3 m/s downward after 2.2 s. To find the distance the camera has fallen after 4.3 s, the gravity acceleration on the Moon must be calculated using the formula g = G.M/R^2. With this information, the formula Vy = Vy0 + a.t can be used to find the final position of the camera, using the initial position of 14.7 m and the final velocity of 3.3 m/s.
lbutscha
During a walk on the Moon, an astronaut accidentally drops his camera over a 14.7 m cliff. It leaves his hands with zero speed, and after 2.2 s it has attained a velocity of 3.3 m/s downward. How far has the camera fallen after 4.3 s?

I am so lost! I have no idea what to do! I would appreciate any help!

First you have to calculate (or search) the gravity acceleration on the Moon:

g = G.M/R^2 (G: gravitational constant, M, R are the mass and radius of the Moon)

Then everything is straight forward.

then would i just multiply that answer by 4.3 for the time?

pixel01 said:
First you have to calculate (or search) the gravity acceleration on the Moon:

g = G.M/R^2 (G: gravitational constant, M, R are the mass and radius of the Moon)

Then everything is straight forward.

With the information he was given, the following formula is better:

http://en.wikipedia.org/wiki/Torricelli's_equation

Where: Vy is the final speed, Vy0 is the initial speed, a is the acceleration, D is displacement. a is the only unknown, so this should be fairly simple.

I am trying to find how far it has fallen. I assume the final speed is the 3.3m/s stated in the problem and the initial is 0m/s. The acceleration on the moon that I figured out from this equation was 1.6256. So should I solve for the displacement? I am so frustrated. I have been working on this since 6 o'clock and I only have half the assignment done.

Woops, I'm sorry, I thought you had the displacement from the first 2.2s...

With the only known information being the variation of speed and time, you should use:

Vy = Vy0 + a.t

Where t is time. With the moon gravity obtained (which must be negative, make sure you get the signs right, and I got -1.5m/s), you have to use another equation:

x = x0 + v0.t + 1/2.a.t^2

Where x0 and x are initial and final position, respectively. You should be able to do this now. Now the exercise asks for the position at 4.3s. Solve for x, and make x0 = 14.7m.

## 1. What is the equation for solving this physics problem?

The equation for solving this physics problem is d = 1/2 * g * t^2, where d is the distance the camera falls, g is the acceleration due to gravity on the moon, and t is the time it takes for the camera to fall.

## 2. How do I determine the acceleration due to gravity on the moon?

The acceleration due to gravity on the moon is approximately 1.62 m/s^2. This value is one-sixth of the acceleration due to gravity on Earth.

## 3. What is the time it takes for the camera to fall?

The time it takes for the camera to fall can be calculated using the equation d = 1/2 * g * t^2, where d is the distance the camera falls and g is the acceleration due to gravity on the moon. Simply rearrange the equation to solve for t.

## 4. Can I use the same equation to solve this problem on Earth?

No, you cannot use the same equation to solve this problem on Earth. The acceleration due to gravity on Earth is different from that on the moon, so you would need to use a different equation.

## 5. Are there any other factors that could affect the outcome of this problem?

Yes, there are other factors that could affect the outcome of this problem. Air resistance and the initial velocity of the camera could also play a role in determining the distance the camera falls. However, for the purposes of this problem, these factors are assumed to be negligible.

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