The question below shows the velocity of three balls projected in different directions will be the same upon hitting the ground.
Why isn't the velocity of the ball thrown horizontally the greatest?
Wouldn't this have both a horizontal velocity and a vertical, downward velocity due to gravity. Hence, wouldn't its velocity be the resultant velocity i.e. $sqrt$ (horizontal velocity$^2$ + vertical velocity$^2$), which would be greater than simply the vertical velocity of ball 1 or 2?
Following from this, the kinetic energy is said to be equal in all 3 when hitting the ground. But wouldn't the horizontally projected ball have the greatest kinetic energy, if it had the greatest speed?
EDIT: After the prompts in the comments, my calculations would go like this:
A̶l̶l̶ ̶t̶h̶r̶e̶e̶ ̶w̶o̶u̶l̶d̶ ̶h̶a̶v̶e̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶v̶e̶r̶t̶i̶c̶a̶l̶ ̶c̶o̶m̶p̶o̶n̶e̶n̶t̶ ̶o̶f̶ ̶v̶e̶l̶o̶c̶i̶t̶y̶ ̶w̶h̶e̶n̶ ̶h̶i̶t̶t̶i̶n̶g̶ ̶t̶h̶e̶ ̶g̶r̶o̶u̶n̶d̶,̶ ̶a̶s̶ ̶t̶h̶e̶i̶r̶ ̶s̶p̶e̶e̶d̶s̶ ̶w̶i̶l̶l̶ ̶a̶l̶l̶ ̶b̶e̶ ̶d̶u̶e̶ ̶t̶o̶ ̶t̶h̶e̶ ̶a̶c̶c̶e̶l̶e̶r̶a̶t̶i̶o̶n̶ ̶f̶r̶o̶m̶ ̶g̶r̶a̶v̶i̶t̶y̶.̶ (Corrected due to comments).
Only ball 3 has a horizontal component of velocity too.
So ball 1 would have velocity $Vy= u + a*t$
Ball 3 would have $Vx=u$ and $Vy=O+a*t$
So the resultant velocity for ball 3 would be $sqrt ( Vx^2 + Vy^2 ) = sqrt (u^2+(at)^2)$ which would be different (less?) than $a*t$.
Is that incorrect?
Where a = acceleration due to gravity = g, t = time, u = initial speed, x = horizontal component and y = vertical component.