Posted by **ruby** on Tuesday, July 22, 2008 at 6:29am.

How would you reword the terms of this equation:

(1/2)M Vo^2 = M g H

Where H is the maximum height above the thrower's hand.

H = Vo²/(2 g) = 11.5 m

Could you substitute the letters H & M with Y, Yo, V, Vo, G, T, or use any of these four newton's equtions:

V = Vo + gt

Y = Yo + Vot + 1/2gt^2

Vo^2 + 2g(Y-Yo)-Y

Average velocity = V + Vo/ 2

Thanx for alll the help. =)

- physics -
**drwls**, Tuesday, July 22, 2008 at 6:34am
No eq

- physics -
**drwls**, Tuesday, July 22, 2008 at 6:35am
No equation was provided

- sorry! -
**ruby**, Tuesday, July 22, 2008 at 6:39am
For this problem:

A person throws a ball upward with an initial velocity of 15 m per second. In order to calculate how high the ball goes, drwls gave me this formula below:

It travels upwards until the vertical velocity component becomes zero. The calculaitons are below:

(1/2)M Vo^2 = M g H

Where H is the maximum height above the

thrower's hand.

H = Vo^2/(2g) = 11.5 m

Could you substitue

H and M for V,Vo,T,Y,Yo,g,a, and could you possible use and of the four newton's equations listed below, to rephrase this?

1.) V = Vo +gt

2.) Y = Yo + Vot + 1/2gt^2

3.) Vo^2 + 2g(Y - Yo)-Y

4.) Average V = V + Vo/ 2

- physics -
**Dr Russ**, Tuesday, July 22, 2008 at 7:05am
For a body already moving upwards with kinetic energy 1/2mu^2 (u=your Vo)the kinetic energy at a point on the upward journey is 1/2mv^2=1/2mu^2-mgh

cancelling m

1/2v^2 = 1/2u^2-gh

or

v^2=u^2-2gh (more familiarly when falling v^2=u^2+2gh)

So for your problem

v^2+2gh=u^2 and v=0 at the top so

2gh=u^2

or h=u^2/2g (=Vo^2/2g as you have)

does this help?

- physics -
**drwls**, Tuesday, July 22, 2008 at 7:54am
I don't know what you mean by "rephrasing" the simple equation

H = Vo^2/(2g)

that provides the answer.

H is the height the ball rises

Vo is the initial velocity that it is thrown upwards

g is the acceleration of gravity (9.8 m/s^2)

If you are looking for another way of deriving the equation starting from Newton's Laws, that can be done, but you will still get the same answer.

By the way, one of your statements

(3) Vo^2 + 2g(Y - Yo)-Y

is not an equation. A correct equation would be

V^2 = Vo^2 - 2 g (Y - Yo)

if Y is measured positive upwards.

This equation leads directly to the one I provided earlier, since Y-Yo = H and V = 0 at the highest point of the trajectory

- physics -
**ruby**, Tuesday, July 22, 2008 at 7:41pm
Yes! thank you both! this really helps1

- physics -
**ruby**, Tuesday, July 22, 2008 at 7:42pm
!* =)

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