Calculating Projectile Height with Position Function | Physics Tutorial

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Homework Help Overview

The discussion revolves around calculating the height of a projectile, specifically a ball thrown upward with an initial velocity and height. Participants are exploring the formulation of the position function that describes height as a function of time.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the possibility of using integration to derive the position function. There are questions about the specific integration process and the equations of motion applicable to the scenario.

Discussion Status

Some participants have provided insights into the integration process needed to derive the position function, mentioning the need to integrate twice and the role of initial conditions in determining constants. Multiple interpretations of the integration approach are being explored.

Contextual Notes

There is an emphasis on the use of constant gravitational acceleration in the discussion, and initial conditions are noted as crucial for determining the constants in the equations.

courtrigrad
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If a ball is thrown upward at 64 ft/sec at an initial height of 80 ft, how would you get the position function that finds the height as a function of t? Do you just integrate?

Thanks :smile:
 
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Of course.Or you may use the equation of motion for a constant gravity field.

Daniel.
 
courtrigrad said:
If a ball is thrown upward at 64 ft/sec at an initial height of 80 ft, how would you get the position function that finds the height as a function of t? Do you just integrate?
What are you proposing to integrate?

AM
 
One would actually integrate TWICE. For this single "z" dimensional problem, one would use:

[tex]v(t) = \int a(t) dt[/tex]

[tex]z(t) = \int v(t) dt[/tex]

where a(t) is vertical acceleration, v(t) vertical velocity, and z(t) vertical displacement. Given constant gravitational acceleration a(t)=g, this would yield:

[tex]v(t) = \int g dt \ = gt + v_0[/tex]

[tex]z(t) = \int (gt + v_0) dt \ = (1/2)gt^2 + v_0t + z_0[/tex]

Initial conditions would determine constants v0 and z0.

~
 

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