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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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