According to the principle of energy conservation, the energy is not created, nor destroyed, it is transformed.
In this problem, we are talking about Mechanical Energy ([tex]M[/tex]) which is the addition of the Kinetic Energy [tex]K[/tex] (energy of the body in motion) and Potential Energy [tex]P[/tex] (It can be Gravitational Potential Energy or Elastic Potential Energy, in this case is the first one):
[tex]M=K+P[/tex] (1)
The Kinetic Energy is: [tex]K=\frac{1}{2}mV^{2}[/tex]
Where [tex]m[/tex] is the mass of the body and [tex]V[/tex] its velocity
And the Potential Energy (Gravitational) is: [tex]P=mgh[/tex]
Where [tex]g[/tex] is the gravitational acceleration and [tex]h[/tex] is the height of the body.
Knowing this, the equation for the Mechanical Energy in this case is:
[tex]M=\frac{1}{2}mV^{2}+mgh[/tex] (2)
Now, according to the Conservation of the Energy Principle, the initial energy [tex]M_{i}[/tex] must be equal to the final energy [tex]M_{f}[/tex]:
[tex]M_{i}=M_{f} [/tex] (3)
[tex]M_{i}=\frac{1}{2}m{V_{i}}^{2}+mgh_{i}[/tex] (4)
At the beginning, the ball is in [tex]h_{i}=0m[/tex] over the point where it is released and has an initial speed [tex]V_{i}[/tex], this means the initial energy [tex]M_{i}[/tex] is only the Kinetic Energy:
[tex]M_{i}=\frac{1}{2}m{V_{i}}^{2}[/tex] (5)
At the maximum height of [tex]h_{f}=19m[/tex] from the point at which the ball was released, it has an speed [tex]V_{f}=0\frac{m}{s}[/tex], because at that very moment the ball stops and then begins to fall.
This means the final energy [tex]M_{f}[/tex] is only the Potential Gravitational Energy
[tex]M_{f}=mgh_{f}[/tex] (6)
Well, according to the explanation above, we have to substitute (5) and (6) in (3):
[tex]\frac{1}{2}m{V_{i}}^{2}=mgh_{f}[/tex] (7)
Now we have to find [tex]V_{i}[/tex], the velocity of the ball when it was released:
[tex]\frac{1}{2}(4.5kg){V_{i}}^{2}=(4.5kg)(9.8\frac{m}{s^2})(19m)[/tex]
[tex]4.5kg{V_{i}}^{2}=2(837.9\frac{kgm^2}{s^2}) [/tex]
[tex]{V_{i}}^{2}=\frac{1675.8\frac{kgm^2}{s^2}}{4.5kg}[/tex]
[tex]{V_{i}}=\sqrt{372.4\frac{m^2}{s^2}}[/tex]
Finally:
[tex]{V_{i}}=19.297\frac{m}{s}[/tex]
