# physics

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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 -

No eq

• physics -

No equation was provided

• sorry! -

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 -

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)

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 -

I don't know what you mean by "rephrasing" the simple equation
H = Vo^2/(2g)
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 -

Yes! thank you both! this really helps1

• physics -

!* =)

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