Motion Question: Speed of 150kg Pile Driver Falling 7.5m

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To determine the speed of a 150-kg pile driver falling from a height of 7.5 meters, the formula v = √(2gh) can be used, where g is the acceleration due to gravity. This equation derives from the conservation of energy principle, equating kinetic energy gained to the potential energy lost during the fall. The calculated speed is approximately 12.1 m/s, which aligns closely with the book's answer of 12 m/s, assuming air resistance is negligible. This approach is effective for free-fall scenarios and can also be used to find the height of an object based on its final velocity. Utilizing energy considerations simplifies the analysis of motion problems like this one.
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Hello,

I have been racking my brains on this one, and I am sure it is an easy question for most. My teacher has been skipping chapters and jumping around in our text a lot and I cannot get the formula to solve this problem. If anyone can help me out with a formula I would be very appreciative. The question is as follows.

A 150-kg pile driver falls from a height of 7.5 m to hit a piling.

What is its spead as it hits the piling.

As I don't have the time it takes to hit the piling, I am unable to tell how fast the pile driver is falling.

I'm hoping this is an easy problem, and in reviewing the chapter I am in, I cannot find a sample problem similar to this, nor a formula that would work. Please help with with a good formula to use to solve this problem.
 
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Try viewing the problem in terms of the energy of the pile driver. The key equation, as in all conservation of energy problems, is

\Delta K = -\Delta U


It is almost always easier to treat "constant acceleration problems" in terms of energy rather then fiddling and twisting the equations of motion in every direction.
 
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Thank you for your response,

My text doesn't seem to cover that forumula in the energy/work chapter. I did run into formula v=Sq Root(2gh) which would be about 12.1 m/s. When checking the answer in the book, it is 12 m/s but is this an accurate formula to use in the future, or a coinsidence in this problems case?
 
That equation will definitely work provided that gravity is the only force, you neglect air resistance. It is derived directly from energy considerations, like quasar recommended.
 
Try deriving it yourself, it follows from the premise that:

The net kinetic energy is a result of a change in potential energy. Algebraically:

\frac{1}{2} mv^2 = mgh

Now solve for v! This works great for freefall (for speeds much less than the escape speed)

Also, solving for h, you can find the height that an object started freefalling from if you know its final velocity.

Energy is very useful when considering such events.
 
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