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

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Gravity on the Moon is significantly less than on Earth, affecting how objects fall. An astronaut's camera dropped from a height of 14.7 m reaches a speed of 3.3 m/s downward after 2.2 seconds, indicating a different gravitational acceleration. To calculate the distance fallen after 4.3 seconds, one must first determine the Moon's gravitational acceleration using the initial and final speeds and time. The height of the cliff is not critical to the calculation, as the camera does not hit the ground within the 4.3 seconds. Understanding these principles is essential for accurately calculating the distance an object falls on the Moon.
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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
 
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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.
 
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