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

[atypical]

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?

 

[Liquidxlax]

\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

 

[atypical]

ok, i understand that it is a velocity, but where is it coming from? why is is the square root of gd/2?

 

[Liquidxlax]

i re read your problem and I'm sorry i gave the wrong advice.

umm v_f² = v_i²+2ad

not sure where the 1/2 came from. sorry its late

 

[atypical]

So from the equation that you posted, they are deriving vi sintheta is greater than or equal to sqart (gd/2)?

 

[atypical]

ok, i understand that it is a velocity, but where is the square root of gd/2 coming from? anyone?