Mechanical energy- Falling rock

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SUMMARY

The discussion focuses on calculating the final speed of a falling rock using the conservation of mechanical energy principle. The original height of the rock is 3.0 meters, and the final velocity just before impact is derived using the equation v2 = sqrt(2g(y2 - y1)). The calculation confirms that the final speed is 7.7 m/s, with y2 correctly taken as zero at ground level. The solution is verified as accurate by other participants in the forum.

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fiziks09
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Homework Statement


If the original height of a falling rock is y1 = h = 3.0m; calculate the rock's speed just before it hits ground.

Homework Equations


From conservation of mechanical energy law

1/2mv1^2 + mgy1 = 1/2mv2^2 + mgy2

The Attempt at a Solution


I think I'm supposed to find v2 i.e final velocity

v1 = 0 (from rest)
y1 = 3
Applying the formula stated above we find that the m's cancel out giving

v2 = sqrt(2g(y2 - y1))
= sqrt(2 x 9.8(3 - 0))
= sqrt(58.8)
= 7.7m/s

Please i want to know if my solution is correct. I'm not sure if it is right to take y2 as zero. Any help will be very much appreciated
 
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Perfectly done!
 
Okay; thanks for verifying !
 

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