The answer is: 8.62m
There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.
Before the ball is thrown, where the potential energy is 0.
After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.
Therefore,
[tex]E=PE+KE[/tex]
Where:
[tex]PE=m.g.h[/tex]
[tex]KE=\frac{1*m*v^{2}}{2}[/tex]
E is the total energy
PE is the potential energy
KE is the kinetic energy
m is the mass of the object
g is the gravitational acceleration
h is the reached height of the object
v is the velocity of the object
Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:
[tex]KE_{1}+PE_{1}=KE_{2}+PE_{2}[/tex]
Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.
[tex]\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}[/tex]
So,
[tex]h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}[/tex]
Hence,
The height at the second moment (maximum height) is 8.62m
Have a nice day!