- #1

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Could you explain how to calculate these sort of things in laymen's terms, or link to a web calculator for these scenarios?

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- Thread starter Researcher X
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- #1

- 93

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Could you explain how to calculate these sort of things in laymen's terms, or link to a web calculator for these scenarios?

- #2

Nabeshin

Science Advisor

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The relevant equations look something like this:

[tex]KE=\frac{1}{2}mv^2[/tex] where m is mass and v is velocity,

[tex]PE=mgh[/tex] Where g is the acceleration due to the earth's gravity near the earth's surface (g=9.8m/s^2 in those units), and h is the height above some fixed point.

For a problem like this, I can say that when I initially throw the ball potential energy is zero and all the energy is kinetic. Then, at the peak, I know it's not moving (kinetic energy is zero) and potential energy is some value associated with height. Using the above definitions of the two types of energy:

[tex]KE_{initial}+PE_{initial}=KE_{final}+PE_{final}[/tex] ,which tells us that the total energy is conserved. From the above argument we can say initial potential is zero, and kinetic final is zero, so we get:

[tex]\frac{1}{2}mv^2=mgh[/tex] Which is another way of saying all the kinetic energy is converted to potential.

Simply rearranging the equation and solving for v:

[tex]v=\sqrt{2gh}[/tex]

The key concept is that the change in energy is zero for any two points along the ball's path. Hope that helps :)

- #3

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I'm not sure what the formula is yet.

- #4

Nabeshin

Science Advisor

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I'm not sure what the formula is yet.

I'm not sure what you're uncomfortable with. Ignoring the derivation, the equation is:

[tex]v=\sqrt{2gh}[/tex]

Where v is the velocity (speed) with which you threw the ball, g is the acceleration due to gravity (9.8 in units of meters per second per second, ~32 in units of feet per second per second), and h is the peak height.

Are you not familiar with the square room symbol?

- #5

- 144

- 3

where,

a = acceleration (in m/s^2)

s = distance (in meters)

v = velocity (in m/s, that’s “meters per second” for clarity purposes)

SQR() = square root of value in parenthesis

9.144 meters = 30 feet

v = SQR(2sa)

SQR(2 * 9.144 meters * 9.8 m/s^2) = 13.38739706 m/s

Note: Where gravity is the acceleration, ‘g’ is commonly used in place of ‘a’ (value of ‘g’ is 9.8 m/s^2 for Earth gravity) and ‘h’ (for height) is commonly used in place of ‘s’ (distance), so the equation becomes:

v = SQR(2hg)

- #6

- 93

- 0

Are you not familiar with the square room symbol?

I know the square root symbol from a calculator where you can put in 64, press it, and get 8.

Being that It's wrapped around the 2gh, I guess v is the square root of 2gh, but what's the "2" for? I've seen 2 come after things and it means squared, or if it's a 3, it means cubed. I'm not sure why the 2 is before both the gravity and the height.

So, if we say the max height was 10 meters, would the velocity be the square root of 10 and 9.8 plus each other?

- #7

Nabeshin

Science Advisor

- 2,205

- 16

I know the square root symbol from a calculator where you can put in 64, press it, and get 8.

Being that It's wrapped around the 2gh, I guess v is the square root of 2gh, but what's the "2" for? I've seen 2 come after things and it means squared, or if it's a 3, it means cubed. I'm not sure why the 2 is before both the gravity and the height.

So, if we say the max height was 10 meters, would the velocity be the square root of 10 and 9.8 plus each other?

The two is just a number multiplying both the 9.8 and the height. It doesn't really have anything to do with the square root. Here, let me give you an example.

Take your original problem of something that goes 30m up.

[tex]v=\sqrt{2*g*h}[/tex]

So, [tex]v=\sqrt{2*9.8*30}=\sqrt{588}\approx 24.5 [/tex] meters per second

Or, again for the example you just used of 10 meters,

[tex]v=\sqrt{2*9.8*10}=\sqrt{196}=14 [/tex] meters per second

That's basically as much as you need to know for using the formula. Hope this makes sense!

- #8

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Perfect! That makes perfect sense now. Thanks.

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