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

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SUMMARY

The discussion focuses on calculating the distance a camera falls after being dropped from a height of 14.7 meters on the Moon. The astronaut drops the camera with an initial velocity of zero, and after 2.2 seconds, it reaches a velocity of 3.3 m/s. The gravitational acceleration on the Moon is derived from the velocity information, leading to the conclusion that the camera falls a specific distance after 4.3 seconds using the formula y = y0 + vy0t - 0.5gt².

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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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