Determining Maximum Height of a Projectile Shot Upward from Earth

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Homework Help Overview

The discussion revolves around determining the maximum height a projectile reaches when shot straight upward from the Earth's surface at a specified speed. The subject area includes concepts from classical mechanics, particularly energy conservation and gravitational potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of energy conservation equations and question the definitions of potential energy and reference points. There are discussions about the applicability of certain formulas, such as mgh, and the implications of using different zero points for potential energy.

Discussion Status

The conversation is ongoing, with participants providing guidance on potential pitfalls in calculations and questioning assumptions made by the original poster. Multiple interpretations of the problem are being explored, particularly regarding the use of gravitational potential energy and the correct setup of equations.

Contextual Notes

There is a mention of the speed being defined in km/hr, and participants note the need for specific values such as the radius of the Earth to proceed with calculations. The discussion also highlights the importance of understanding the limits of certain equations based on the height of the projectile relative to the Earth's radius.

splac6996
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Homework Statement


A projectile is shot straight up from the Earth's surface at a speed of 1.40×10^4 . How high does it go?



Homework Equations


potential energy is given by (GM(1)M(2))/R


The Attempt at a Solution


I attempted to solve this using energy conservation equation and solving for r which I thought would be the radius of the Earth plus the distance traveled by the projectile. By subtracting the radius I thought I would get the distance traveled but my answer is wrong can someone help please
 
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splac6996 said:
A projectile is shot straight up from the Earth's surface at a speed of 1.40×10^4 . How high does it go?

What units is that speed in?

It might be of help to a reader who is going to assist you if you would show the actual calculation you did.
 
There are a number of things you could have done wrong. For example, did you remember to use a negative sign in your potential? Do you have an initial potential, where did you define your zero in potential energy to be? Specifically, what was the equation you used?
 
the speed is defined in km/hr and for the formula that i used I did use a negative sign for my potential energy and I used my zero point to be the point were the projectile is shot from.
 
Why not just use m.g.h=1/2mv^2?

And potential energy is in Joule not in Newton.
 
katchum said:
Why not just use m.g.h=1/2mv^2?

And potential energy is in Joule not in Newton.

mgh is only applicable near the Earth's surface. For this projectile it will be traveling significantly far away from the Earth's surface to make mgh non-applicable.
 
Well my understanding is that if the distance is smaller than than the radius then I could use mgh=1/2mv^2 but in this case since the height is going to be largerthan the radius of the Earth so my answer would be wrong.
 
use v^2=u^2+2as,

your deceleration is not constant in the equation. The deceleration is given as the average deceleration the object undergo in that motion. to find the average deceleration,

integrate the below equation with respect to distance, with the radius of the Earth as the lower limit and the radius+height of the Earth as the upper limit.

then divide by the height to find the average deceleration.

a=Gm/r^2

EDIT: Well, you will need to have the radius of the Earth to solve the equation. Even after simplifying expressions of the radius of the Earth into g.
 
Last edited:
@splac6996: Well, If you take zero of your PE at surface of the Earth, your formula for PE at R (in #1 post) is not correct.
 
Last edited:
  • #10
this is what i think
-GM(earth)m(projectile)/H(radius of Earth + distance traveled)=(1/2)m(projectile)v^2
doing the algebra and simplifying i have
H=-2GM/v^2
is that correct
 
  • #11
Nope. You can see, if you put distance traveled = 0 your
RHS = 0 but LHS is not = 0.
 
  • #12
the expression for the potential -GM(earth)m(projectile)/H(radius of Earth + distance traveled) gives you the potential energy at the highest point. You will still need to minus the potential at the Earth's radius.
 
  • #13
With gravitational force [itex]F= -GMm/r^2[/itex], the potential energy is [itex]GMm/r[/itex] but that has 0 point at infinity, not at the surface of the earth.
 
  • #14
Can you show the exact figures you got?
 

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