Question

PLEASE HELP! I DON'T KNOW THIS AND IT IS A BIG GRADE.

The distance traveled by a falling object is given by the formula d = 0.5gt2 where d = distance, g = the acceleration due to gravity, and t = time. Solve this equation for g, and use your formula to determine the acceleration due to gravity if a baseball takes 10 seconds to hit the ground after being dropped from a height of 490 meters.

Answer 1

g = 9.8 ft/sec² ANSWER

Answer 2

thank you so i would just right g=9.8 feet/seconds? @lionblood

Answer 3

This might help you. http://openstudy.com/study#/updates/530bb250e4b04a7f3c1db864

Answer 4

that one says feet,mine is meters @lionblood

Answer 5

Hit a dilemma? ^^

Answer 6

yes i have no idea how to do this @Supreme_Kurt

Answer 7

Just call me Kurt ^^
Now, the reason you're getting all lost and confused is because you weren't showed how to actually solve this problem, like, step-by-step.

If you'll let me, I'll be able to do just that :)

Ready?

Answer 8

yes thank you

Answer 9

Now, start with this:
\[\Large d = 0.5\color{red}gt^2\]
And we need to figure out the "acceleration due to gravity" or g.

To do that, we need to get g ALL ALONE on one side of the equation.

Any idea how to proceed with that? :)

Answer 10

multiply the sides by 0.5?

Answer 11

Almost right.
In time, you'll learn that "almost right" is just a nice way of saying "wrong"

But that's okay. Better you make mistakes here than in the actual exams or something. This is how we learn :D

If I'm not mistaken, your goal is to get rid of the 0.5 on the right-side, yes?
Your heart's in the right place (hehe)
But multiplying both sides by 0.5 won't do the trick.
That'll just turn the 0.5 on the right-side into a 0.25.


If that 0.5 were a 1 instead, that'd be as if it just disappeared, since 1 multiplied to anything gets you, well, that thing. You know what I mean.

That said, what do you multiply to 0.5 so that you get 1? It sure as hell isn't 0.5 ;)

Answer 12

0.5x2 would be 1

Answer 13

That is correct :D
Now, if we multiply both sides by 2, what happens to the equation? ^^

Show me.

Answer 14

i dont know :( this makes me look stupid

Answer 15

Hey, just to put this there, if you're stuck or you think you've hit a dead end, tell me immediately. I'm here to teach you, not judge you ^^

And you're not stupid unless you admit it to yourself, so do yourself a favour and don't do that, okay?

Now, back to our equation.
\[\Large d = 0.5\color{red}gt^2\]

Answer 16

We multiply both sides by 2... nice and slowly...
\[\Large \color{green}2\cdot d = \color{green}2\cdot 0.5\color{red}gt^2\]
With me so far?

Answer 17

so far yes

Answer 18

Okay. 2d. Nothing we can do about that now.
\[\Large \color{blue}{2d} = \color{green}2\cdot 0.5\color{red}gt^2\]
Still with me?

Answer 19

yep

Answer 20

Now, about this part.
\[\Large 2d = \color{blue}{2\cdot 0.5}gt^2\]
2 times 0.5

What does this become?

You said yourself just a while ago ;)

Answer 21

1

Answer 22

That's good. And you know it's pointless to write "1" times anything right, because it doesn't really do anything?

ie

1 times 5 is just 5, so might as well just write 5 instead of 1x5
etc.

Getting me? :)

Answer 23

ok

Answer 24

Okay, I'll demonstrate. :D
\[\Large 2d = \color{blue}{2\cdot0.5}gt^2\]
2 times 0.5 is just 1.
\[\Large 2d = \color{blue}1gt^2\]
But 1 times gt^2 is jugt gt^2 so...
\[\Large 2d = gt^2\]

okay? ^^

Answer 25

okay

Answer 26

Now, we need to get rid of that t^2.

It's multiplied to g, so to remove it, you...?

Answer 27

multiply?

Answer 28

No... okay, to undo multiplication, you divide :D
To undo division, you multiply.
To undo addition, you subtract.
To undo subtraction, you add.


This is basic stuff, so please remember it :D

To undo the multiplication of t^2, you *divide* both sides by t^2.
And you get...?

Answer 29

this is way to hard for me, i wanna say g^2?

Answer 30

No... okay, let me show you :D
\[\large 2d = gt^2\]
We now divide both sides by t^2
\[\Large \frac{2d}{t^2}= \frac{gt^2}{t^2}\]
What happens next?

Answer 31

you solve it

Answer 32

Of course -_-

But specifically, what step comes next? :)

Answer 33

\[\Large \frac{2d}{t^2}=\frac{g\color{blue}{t^2}}{\color{red}{t^2}}\]

These two cancel out, don't they?

Answer 34

ohh yes

Answer 35

And what does that leave you with?

Answer 36

2d over t^2 = gt?

Answer 37

Why would there be a t left over? :P

Answer 38

ohh sorry lol, 2d over t^2 = g

Answer 39

And that's it, you've solved for g, as per the instructions of the problem.
\[\Large g = \frac{2d}{t^2}\]

Answer 40

Now, to solve for the actual value of g, well, we need d and t.
distance and time.

Any idea what those are? :)

Answer 41

no :(

Answer 42

4?

Answer 43

Two numbers are staring you right in the face.
I quote your problem:
"...if a baseball takes 10 seconds to hit the ground after being dropped from a height of 490 meters. "

The *distance* and the *time* must be there somewhere...

Answer 44

490 is distance and 10 seconds is the time

Answer 45

That's correct. Now just plug those in. d is distance, t is time:
\[\Large g = \frac{2(490)}{(10)^2}\]
and do the math :D

Answer 46

g=980 over 100?

Answer 47

Yup. Simplify? :)

Answer 48

9 seconds?

Answer 49

nope. Too simple.
Keep it precise.
to divide by 100, simply move the decimal point two places to the left. ;)

Answer 50

what would it be, im stuck

Answer 51

@Supreme_Kurt

Answer 52

980.0
move the decimal point two places to the left... to divide it by 100.

and you get...?

Answer 53

9.8?

Answer 54

Which is indeed, what Lion gave you (sort of)

Except now, you know where it came from :)

It's m/s^2 btw, not feet/s^2.

That's it :D

Answer 55

thank you so much :)

Answer 56

No problem :)

Practice makes perfect, so keep right on practicing, okay? ^^