What Is the Initial Speed and Maximum Height of a Rock Thrown Vertically?

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of a rock thrown vertically into the air. The original poster provides specific parameters, including the weight of the rock, its height, and speed at that height, and seeks to determine the initial speed and maximum height using the work-energy theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of both kinematic equations and the work-energy theorem to find the initial speed and maximum height. There are discussions about the relationship between initial and final energies, and the application of gravitational work.

Discussion Status

Participants are actively engaging with the problem, raising questions about the application of energy concepts and the calculations involved. Some guidance has been offered regarding the relationship between kinetic and potential energy, but there is no explicit consensus on the next steps or methods to proceed.

Contextual Notes

Participants are navigating the constraints of the problem, including the absence of external forces other than gravity and the need to calculate work done by gravity. There are also discussions about the definitions of work and energy in this context.

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Homework Statement



You throw a rock of weight 20.4 N vertically into the air from ground level. You observe that when it is a height 14.9 m above the ground, it is traveling at a speed of 26.0 m/s upward.

Use the work-energy theorem to find its speed just as it left the ground;
Use the work-energy theorem to find its maximum height.
Take the free fall acceleration to be g = 9.80 m/s^2.


Homework Equations


Wtotal = delta K

K = mv^2

v^2 = vi^2 + 2a(x-xi)

The Attempt at a Solution




I believe this can also be done using kinematic equation, since we are given vf,x,a, and we want vi.

So if this is true, I will attempt both ways :D


26^2 = vi^2 + 2(-9.8)(14.9-0)
vi = 31.11 <--------> which is right o:)

now, for using the work energy theorem.

we first need the mass.

using
w=mg

20.4 = m ( 9.8)

m = 2.08

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2

K2 = 843.65
K1 = 0

...how do I proceed?
 
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since there is no external force (such as friction or air resistance) there is no external work done on theb all except for gravity. So the total initial energy of the ball is equal to the total final energy of the ball

what is the initial energy of the ball and the final energy of the ball?
 
ummm...

so Winitial = Wfinal.

W=Fs

W = (F)(14.9)...?
 
Heat said:
ummm...

so Winitial = Wfinal.

W=Fs

W = (F)(14.9)...?

No, it's not "work initial" = "work final", which in itself would not make sense.

you may either use

change of kinetic energy = total work done by gravity

OR

total final energy = total initial energy

where total energy = kinetic energy plus gravitational potential energy
 
delta k = wg ?

if so, how do I apply this to my problem, as I have this:

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2
 
Heat said:
delta k = wg ?

if so, how do I apply this to my problem, as I have this:

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2
ok so u have the final and initial kinetic energies
now how do you calculate the work done by gravity??
 
w done by gravity is:

w = mgsin(theta)

w = 2.08 (9.8)(sin 90)
w = 20.38
 

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