# Position in Time

1. Feb 9, 2005

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

2. Feb 9, 2005

### dextercioby

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

Daniel.

3. Feb 10, 2005

### Andrew Mason

What are you proposing to integrate?

AM

4. Feb 10, 2005

### xanthym

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

$$v(t) = \int a(t) dt$$

$$z(t) = \int v(t) dt$$

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:

$$v(t) = \int g dt \ = gt + v_0$$

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

Initial conditions would determine constants v0 and z0.

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