Answer:
[tex]v_{0}=42.23m/s[/tex]
Explanation:
We shall use conservation of energy to solve this problem
Initial energy of the ball = Kinetic energy + potential energy
[tex]Energy_{Initial}=\frac{1}{2}mv_{0}^{2}+mgh[/tex]
When the ball finally hit's ground it's potential energy becomes zero thus final energy becomes
[tex]Energy_{final}=\frac{1}{2}mv_{f}^{2}+0\\\\Energy_{final}=\frac{1}{2}m(1.2v_{0})^{2}\\\\Energy_{final}=0.72mv_{0}^{2}[/tex]
Equating both the energies and applying the given values we get
[tex]\\\\0.5mv_{0}^{2}+mgh=0.72mv_{0}^{2}\\\\0.22v_{0}^{2}=gh\\\\v_{0}=\sqrt{\frac{g\times 40}{0.22}}\\\\v_{0}=42.23m/s[/tex]
