Help please, calculation involving height, speed and gravity

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Homework Help Overview

The discussion revolves around a calculation involving height, speed, and gravity, specifically focusing on the conservation of energy principles in a physics context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between mass and energy calculations, questioning whether mass can be omitted in certain equations. They discuss the conservation of energy and its implications for kinetic and potential energy.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting that mass can be treated as a variable that cancels out in the equations. There is an ongoing exploration of the implications of this approach, with participants clarifying the role of mass in energy equations.

Contextual Notes

There is a mention of kinematic equations and specific variables such as initial velocity and final displacement, indicating that certain information may be assumed or provided for the calculations.

DB10
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Solved. Thanks.
 
Last edited:
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You can actually do the problem without knowing the mass of the ball. Instead of using 155g, do all of the calculations using 'm', and you should find that it will cancel out in the process of the calculation.

Consider the law of conservation of energy -- the sum of the kinetic and potential energies at the instant the ball is thrown should be equal to the sum of energies at the instant the ball hits the ground.
 
So I can use the two above equations but not include the 'm'? Thanks for your help.
 
Well you would still include the m, but they will eventually cancel out, so omitting the m right from the start will not make a difference (although it is probably a bad habit to cancel things out before it actually happens during the process of the calculation).

What i mean is that initially, as the ball is released, it has kinetic and potential energy (with reference to the ground) given by:

[tex]E_k = 0.5 mv_1^2[/tex]
[tex]E_p = mgh[/tex]

And as it hits the ground:

[tex] E_k = 0.5mv_2^2[/tex]
[tex]E_p = 0[/tex]

Once you create an equation from these terms, you will see that every term has an 'm' in it which can be eliminated by dividing everything by m.
 
Consult the Kinematic Equations (see attached).

You know:
The initial velocity, Viy
The final displacement, Dy
The acceleration, a

You are looking for the final velocity, Vfy
 

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