How Is Gravity Calculated for a Dropped Object on the Moon?

[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?

The Attempt at a Solution

y= y0+vyot-.5gt^2
y=14,7-.5(9.8) 4.3^2
My answer was not correct and I have no idea where to begin again! Please help

 

[Shooting Star]

First find the g on the moon by using the info that its speed is 3.3 m/s after 2.2 s. Do you know the formula connencting initial and final speeds, accn and time?

Then find the dist covered in 4.3 s by using the formula you've written above.

The height of the cliff plays no major role in all this, except to tell us that the camera has not crashed into the ground before 4.3 s.

 

[Moetasim]

!

I would approach this problem by first understanding the basic principles of motion and gravity. The equation used, y=y0+vyot-.5gt^2, is the formula for displacement (y) which includes the initial position (y0), initial velocity (vo), acceleration due to gravity (g), and time (t).

In this scenario, the initial position (y0) is 14.7m, the initial velocity (vo) is 0 m/s (since the camera is dropped with zero speed), and the acceleration due to gravity (g) is 9.8 m/s^2 (this value is a constant on the surface of the moon). We are trying to find the displacement (y) after 4.3 seconds, so we can plug those values into the equation as follows:

y= 14.7 + (0)(4.3) - 0.5(9.8)(4.3)^2
y= 14.7 - 0 - 0.5(9.8)(18.49)
y= 14.7 - 90.706
y= -75.006

Therefore, after 4.3 seconds, the camera has fallen 75.006 meters downward. It is important to note that the negative sign indicates the direction of the displacement, which in this case is downward.

I would also suggest double-checking your calculations and units to ensure accuracy. Additionally, it is always helpful to draw a diagram or visualize the scenario to better understand the problem and the solution.