Velocity Calculation Contradiction

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SUMMARY

The discussion centers on a physics experiment involving the calculation of final vertical velocity (Vfy) after projecting an object horizontally from a height of 2 meters. The participant calculated Vfy to be 6.261 m/s using the equation Vfy^2 = Viy^2 + 2ad, where acceleration (a) is 9.8 m/s². However, confusion arose when calculating the time interval (t) using the derived Vfy, leading to a horizontal velocity (V) of 3.13 m/s, which contradicted the initial findings. The participant concluded that the first calculation for Vfy should be prioritized due to the assumption of constant velocity in the horizontal motion equation.

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Homework Statement


I have to design an experiment where I am supposed to project an object horizontally from some point above ground level, measure the distances, calculate the average horizontal distance dx, then use that to calculate the initial and final velocities yatta yatta.
Anyway I made the height 2 m, got 4.62 m as the average horizontal distance and I want to get Vfy (y component of final velocity)

Homework Equations


Vf^2 = Vi^2 + 2ad
a = (delta)V/t
V = d/t
where: Vf=final velocity, Vi=initial velocity, a=acceleration(9.8 m/s^2), t=time interval, d= distance

The Attempt at a Solution


I can't complete the calculations in the experiment because I can't get one value for Vfy

Vfy^2= Viy^2 + 2ad
Vfy = (sqrt)[0 + 2(9.8m/s2)(2m)]
= 6.261 m/s
great, now to find time:
a = (delta)V/t
t = (delta)V/a
= (6.261m/s)/(9.8 m/s2)
= 0.639 s

but when I use the time in this equation:
V = d/t
= (2 m)/(0.639s)
= 3.13 m/s
WHAT?! The same velocity is now half as small. How did this happen? Which one is the right value?
Something tells me I should just go along with the first one.

Oh by the way, Hi! :)
 
Last edited:
Physics news on Phys.org
V = d/t is only valid when velocity is constant.
 
Lol I keep forgetting things like acceleration. Thanks btw :)
 

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