# 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.

Note by Xuming Liang
5 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 5 years, 1 month 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

- 5 years, 1 month ago