Answer:
[tex]H=5.33\ m[/tex]
Step-by-step explanation:
Conservation of the Energy
The total mechanical energy of an object is conserved if no external forces (like friction) are applied. When the ball is thrown upwards, the gravitatory potential energy is 0 because its height is also 0. But the kinetic energy is at max because the speed is at its maximum value. Let's assume vo is the initial speed and H is the maximum height the ball will eventually reach. The initial kinetic energy is totally converted to gravitational potential energy when the ball reaches its maximum height and the speed is zero, thus
[tex]\displaystyle mgH=\frac{mv_o^2}{2}[/tex]
Solving for H
[tex]\displaystyle H=\frac{v_o^2}{2g}[/tex]
To find vo, we use the given condition: at h=4 m above the launch point the speed is one-half the launch speed. Thus the total energy at that moment is
[tex]\displaystyle E=mgh+\frac{m(\frac{v_o}{2})^2}{2}[/tex]
Operating
[tex]\displaystyle E=mgh+\frac{mv_o^2}{8}[/tex]
This total energy is equal to the initial kinetic energy:
[tex]\displaystyle mgh+\frac{mv_o^2}{8}=\frac{mv_o^2}{2}[/tex]
Simplifying by m and multiplying by 8
[tex]\displaystyle 8gh+v_o^2=4v_o^2[/tex]
Solving for vo
[tex]\displaystyle v_o^2=\frac{8gh}{3}[/tex]
Finally, replacing in the equation for H
[tex]\displaystyle H=\frac{\frac{8gh}{3}}{2g}=\frac{4h}{3}=\frac{4\cdot 4}{3}=5.33\ m[/tex]
[tex]\boxed{H=5.33\ m}[/tex]
