Finding the Speed of a Falling Ball with Constant Acceleration

AI Thread Summary
To determine the speed of a falling ball after 1.25m in 0.5s with constant acceleration, the correct approach involves using the equations for uniformly accelerated motion. The formula s = (1/2)at² + v₀t applies, where s is the distance, a is acceleration, v₀ is initial velocity (zero in this case), and t is time. The final speed can also be calculated using v = at + v₀. The confusion about the factor of 2 in the initial calculation stems from misunderstanding the average speed during the fall. Combining these equations will yield the correct final speed of the ball.
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Homework Statement



A small steel ball, dropped from rest, takes 0.5s to fall through a vertical distance of 1.25m.
Determine the actual speed of the ball when it has fallen through a vertical distance of 1.25m, assuming that acceleration is constant.


Homework Equations



v=s/t

The Attempt at a Solution



Long ago I was helped by a friend to get the result
( 1.25m x 2 ) / 0.5s = 5ms-1

but I've forgotten how did he come to that conclusion, and the reason behind (1.25mx2), help is much appreciated .
 
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The formula s=vt is only valid for constant speed. Does an object fall with constant speed?

Your friend used the following, correct equations:

1) s=\frac{1}{2}at^2+v_0t+x_0
2) v=at+v_0

These are valid for constant acceleration. You will need to combine both of them to get the answer.
 
s = t*(u+v)/2
 
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