Finding the position function for a ball thrown upward

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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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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.

~