So, V_F= 11m/sWhat is the speed of the potted plant after falling 6m?

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SUMMARY

The speed of the potted plant after falling 6 meters from a 12-meter high ledge is calculated to be approximately 11 m/s. This conclusion is derived using the principles of kinetic and potential energy, specifically the equation v² = 2gh, where g is the acceleration due to gravity (9.81 m/s²) and h is the height (6 m). An alternative method using the formula for final velocity, v_f = √(2AD), also yields a similar result of 10.844 m/s. Both methods confirm that mass is irrelevant in this scenario due to the conservation of mechanical energy.

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Homework Statement


[PLAIN]http://img8.imageshack.us/img8/664/photo4dc.jpg
Sorry it is sideways.

Only known data is the 6m drop from the 12m high ledge.

Homework Equations



Currently wokring with Kinetic energy (1/2mv^2), potential energy(mgh).

The Attempt at a Solution



My thinking is mass doesn't matter, it will cancel out due to the mechanical energy( if that makes sense). Therefore the equation to solve is v2f=2gh

Therefore after subbing in 9.81m/ss for g, 6m for h; my final answer is 11m/s with rounding because of sig figs.

Thank you for help
 
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Uh, do you think you could make that figure a little bigger ... I mean, I can almost see 25% of it at a time as it is.
 
I also got 11m/s, but found that answer in a less complicated way:

You know:
velocity_initial is 0m/s
Distance is 6m
Acceleration is 9.8m/ss

So:
velocity_final= SQRT(2*A*D)
V_F= SQRT(2*9.8*6)
V_F= 10.844m/s
 

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