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. =)
No eq
No equation was provided
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
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?
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
Yes! thank you both! this really helps1
To reword the terms in the equation (1/2)MV0^2 = MgH, you can use the following substitutions:
1. Substitute H with Y: (1/2)MV0^2 = MgY
2. Substitute M with T: (1/2)TV0^2 = TgY (T is a new symbol representing mass)
3. Substitute Y with X: (1/2)TV0^2 = TgX
Now, for the given values H = 11.5 m, you can substitute it into the equation to solve for the other variables:
(1/2)TV0^2 = Tg(11.5)
Next, you mentioned four equations formulated by Newton. Let's analyze if they can be used to substitute into the equation:
1. V = V0 + gt: This equation can help you find the final velocity (V) if you have the initial velocity (V0), acceleration due to gravity (g), and time (t). It does not directly relate to the given equation.
2. Y = Yo + V0t + 1/2gt^2: This equation can help you find the displacement (Y) if you have the initial position (Yo), initial velocity (V0), time (t), and acceleration due to gravity (g). It does not directly relate to the given equation.
3. V0^2 + 2g(Y - Yo) - Y: This equation relates to the given equation, but it rearranges the terms in a different way. You can use it to solve for V0^2.
4. Average velocity = V + V0/2: This equation relates to the given equation indirectly, but it does not directly help to substitute any terms.
To summarize, you can substitute Y = 11.5 into the equation (1/2)TV0^2 = Tg(11.5) and solve for the other variables. Additionally, you can use Newton's third equation, V0^2 + 2g(Y - Yo) - Y, to find V0^2 if you have the other variables.