physics...no friction??????

In (somewhat!) simpler terms,

V = √(2gy) = √(2*9.8*3.0) = 7.668 m/sec

You mean "how FAST is it traveling".

PE = KE; potential energy = kinetic energy because of the conservation of energy gig. Thus, PE = mgh = 1/2 mv^2 = KE and v = sqrt(2gh); where g = 9.81 m/sec^2 at Earth's surface, h = 3 m = height above the hand as the ball leaves it, and v = velocity as the ball leaves the hand. You can do the math.

Physics lesson learned: mass is meaningless in this because it cancels out. THIS IS IMPORTANT because it is consistent with Newton's observations that two masses (M and m), when released from the same height (h), will hit the ground at the same time. That is, their accelerations and average velocities are the same regardless of the mass.

PS: Notice, there is no need to assume "no height" for the tosser. What we are interested in is the KE that is converted into PE at the height above the hand (where the KE is) when the ball is let go.

Why? Because it is the KE at the hand when the ball is released, not the KE at ground level, that is converted into PE at a height = 3 m above the hand. Clearly KE on the ground, if the ball does in fact go all the way to the ground, will be greater than the KE as the ball leaves the hand.

Point is that PE = KE if and only if we assume the height going up is equal to the height it falls. In your problem, height is measured from the hand position when the ball is released, not the ground.

I assume you mean how fast it is traveling, as how far it traveled is given. kinetic energy, Ek=m(V^2)/2, while potential gravitational energy Ug=mgh. Since energy is conserved in this system, we have: m(v^2)/2=mgh => V^2=2gh => V=sqrt(2gh)=sqrt(2*3*10)[this is using the easier gravitational constant g=10 instead of 9.8]

This answer assumes the ball was thrown straight up and that the thrower is of no height.

Vf = final velocity (will be zero because at the peak of the throw, velocity is zero)

Vi = initial velocity (the unknown we're solving for)

g = gravity (-9.8 m/s)

y = the height the ball is thrown

you don't need weight for this method

Vf^2 = Vi^2 + 2gy

plug in and solve for your answer

how far? 3 meters