Explanation for an easy(not for me) physics question

I don't think this would've been a problem on Brilliant since I found it in a physics workbook, the problem is: A boy stands on a 50-m-high bridge with his GI Joe. The action figure has a real parachute. The boy drops Joe from the bridge, but tragically the chute fails to deploy. If the action figure has a mass of 100 g and it never reaches terminal velocity, how fast will Joe be going when he strikes the river?

After reading this question, at first I was confused about why the mass of the object is given. Since the object experiences a constant gravitational pull from earth regardless of its mass, couldn't we just use the formula 50=at2250=\frac{at^2}{2} to find the time then multiply it by a(or 9.8)? Surprisingly, the answer given in the back of the book used the conservation of energy(with a different answer also),which I would never have thought of.Why??? Do I use conservation of energy in a problem whenever the mass of an object is given?(doesn't make any sense...)

Second question which is not about the problem above: It seems like a lot of ideas in physics are really abstract. I managed to understand and visualize force and work, but not momentum and power yet. How do you guys overcome this problem? I would like to know.

Sorry for my ignorance :(. Thank you.

Note by Xuming Liang
6 years, 3 months ago

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4 votes

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There's nothing wrong with your approach. A slightly shorter way is this:


where, vv is Joe's final velocity, gg is the gravitational acceleration and hh is the height Joe is falling from.

If we plug in the values we get v=2×9.8×50ms1v=\sqrt{2\times9.8\times50} ms^{-1} or approximately 31.3ms131.3 ms^{-1}.

We don't need the mass to figure this out. We can use the law of conservation of energy, but it'll give us the same result. The mass terms will cancel out. I don't know why your book has a different answer.

Mursalin Habib - 6 years, 3 months ago

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try to read paul g,hewitt concepts of physics it 'll help and comin 2 the 1st one u must get the same answer in both ways

Naga Teja - 6 years, 3 months ago

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