Where Does the Square Root of gd/2 Come From?

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SUMMARY

The discussion centers on the derivation of the equation \(\sqrt{\frac{gd}{2}}\), which represents the final velocity of an object in free fall from a height \(d\) under the influence of gravity \(g\). Participants clarify that this equation arises from the kinematic equation \(vf^2 = vi^2 + 2ad\), where \(a\) is the acceleration due to gravity. The confusion lies in understanding the relationship between initial velocity components and the derived velocity at maximum height, specifically how the factor of 1/2 is integrated into the equation. Ultimately, the square root of \(gd/2\) is a critical component in analyzing projectile motion.

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  • Understanding of kinematic equations, particularly \(vf^2 = vi^2 + 2ad\)
  • Basic knowledge of projectile motion and components of velocity
  • Familiarity with gravitational acceleration (g) and its effects on falling objects
  • Concept of maximum height in projectile trajectories
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  • Explore the relationship between initial velocity and final velocity in free fall scenarios
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Homework Statement


See picture please

Homework Equations


The Attempt at a Solution


Here is what I understand:
vi sin(theta) = initial velocity height on velocity/time graph
What I don't understand:
sqrt(gd/2) is this gravity times initial distance height of target?
Where is the second side of the equation is coming from?
 

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\sqrt{\frac{gd}{2}}= some velocity that is less than or equal to vsintheta which is another velocity. So when d is at the maximum height both sides should equal each other
 
ok, i understand that it is a velocity, but where is it coming from? why is is the square root of gd/2?
 
i re read your problem and I'm sorry i gave the wrong advice.

umm vf2 = vi2+2ad

not sure where the 1/2 came from. sorry its late
 
So from the equation that you posted, they are deriving vi sintheta is greater than or equal to sqart (gd/2)?
 
ok, i understand that it is a velocity, but where is the square root of gd/2 coming from? anyone?
 

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