Calculating Drag (having circular reference problems)

AI Thread Summary
The discussion revolves around calculating drag for a physics experiment involving a square open-sided cube dropped from a height. The user struggles with circular references in determining displacement and velocity, essential for calculating drag. They provide specific values for parameters like density (p = 1.2), drag coefficient (Cd = 1.05), and area (A = 0.0225 m²). The user calculates the weight of the cube and attempts to find terminal velocity, but encounters issues with negative values in their velocity calculations. The confusion stems from misapplying the kinematic equation, highlighting the importance of correctly interpreting the signs in physics equations.
josiahseto
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Homework Statement



I don't know if it's possible but I am doing a physics EEI on drag.
I know to find drag you need instantaneous velocity which can be found through (change in x/ change in t) but i cannot find displacement unless i find velocity :S..

Homework Equations


d = 0.5 x p x (V^2) x Cd x A
im taking p = 1.2
Cd = 1.05 (for square shape)
A = 0.0225 sq meters

The Attempt at a Solution


I have dropped a square open sided cube 15cm x 15cm through a vertical distance of 2.5m (u = 0m/s) and it took an average of 2.33s
 
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also, its mass is 0.0029 kg
so w = 0.02842N

therefore at terminal velocity point drag = 0.02842N (thats if it does reach TV)

therefore at that point acceleration = 0

so.. we have

u = 0m/s

v = 1.41m/s (calculated through Fd = Fg)

and a = 0m/s at TV point
 
the velocity can be found using v2=u2-2gy
 
v^2 = 0^2 - 2 x 9.8 x 2.5
V^2 = -49
v = sqrt -49 = impossible?
 
josiahseto said:
v^2 = 0^2 - 2 x 9.8 x 2.5
V^2 = -49
v = sqrt -49 = impossible?

Sorry, down is positive, so it would be v^2=0^2+2*9.81*2.5.
 
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