Ball bearing launched from point

AI Thread Summary
To find the initial velocity of a ball bearing launched horizontally from a height of 85.4 cm, the average distance traveled was recorded as 250.24 cm. The calculation involves using the equations of motion, specifically d = vt for horizontal distance and h = 0.5at² for vertical drop. The derived formula v = d sqrt(g / 2h) was confirmed to yield the same result as using the standard horizontal and vertical motion equations. The calculated initial velocity was approximately 5.99 m/s. Understanding these principles allows for calculating both the initial velocity and the distance traveled even without a provided distance.
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Homework Statement


In my physics class we shot a ball bearing horizontally 85.4cm off the ground and we recorded the distance at which it landed. How can I find the initial velocity? The average distance traveled was 250.24 cm.

What if I was not provided I distance, how could I calculate the distance this ball would travel?

Homework Equations


I'm not sure.


The Attempt at a Solution


I have tried doing:

d sqrt(g / 2h)

2.5024(converted to m) sqrt( -9.8 / 2 x -85.4)
= 5.99 m/s

Is this right?
 
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I don't recognize the formula d sqrt(g / 2h) but I get the same answer when I do it with good old horizontal (d=vt) and vertical (d = .5at²) parts.
 
Delphi51 said:
I don't recognize the formula d sqrt(g / 2h) but I get the same answer when I do it with good old horizontal (d=vt) and vertical (d = .5at²) parts.

They're the same:

(1) d = vt

(2) t = \frac{d}{v}


(3) h = \frac{a}{2}t^2

substitute t from equation (2) above:

(4) h = \frac{a}{2} \left(\frac{d}{v} \right)^2 = \frac{ad^2}{2v^2}

(5) v^2 = \frac{ad^2}{2h} = d^2 \left( \frac{a}{2h} \right)

(6) v = \sqrt{d^2 \left( \frac{a}{2h} \right)} = d \sqrt{\frac{a}{2h}}
 
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