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 $50=\frac{at^2}{2}$ 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.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

$v=\sqrt{2gh}$

where, $v$ is Joe's final velocity, $g$ is the gravitational acceleration and $h$ is the height Joe is falling from.

If we plug in the values we get $v=\sqrt{2\times9.8\times50} ms^{-1}$ or approximately $31.3 ms^{-1}$ .

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 - 7 years, 11 months ago

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 - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$