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

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The discussion revolves around the derivation of the equation sqrt(gd/2) in the context of projectile motion. Participants are trying to understand how this expression relates to the initial velocity and the height of a target. The equation is linked to the kinematic formula vf² = vi² + 2ad, where 'g' represents gravity and 'd' is the distance. There is confusion about the origin of the 1/2 factor in the equation and its implications for maximum height. Clarification is sought on how the square root of gd/2 represents a velocity that is compared to vi sin(theta).
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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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