Given,
The initial height of the ball, h₁=6 ft
The initial velocity of the ball, u=64 ft/s
The acceleration due to gravity, g=-32 ft/s²
The velocity of the ball when it reaches the maximum height will be, v=0 ft/s
From the equation of motion,
![v^2-u^2=2gh_2](https://img.qammunity.org/2023/formulas/physics/high-school/4rypb3fakl22jlq9hmy0wgx0j0lwlj84pd.png)
Where h₂ is the height covered by the ball to reach the maximum height.
On substituting the known values,
![\begin{gathered} 0-64^2=2*-32* h_2 \\ h_2=(-64^2)/(2*-32) \\ =64\text{ ft} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/jsmb75pe464ump0a235tbvya2aw7rdap7i.png)
Thus the maximum height reached by the ball is
![\begin{gathered} H=h_1+h_2 \\ =6+64 \\ =70\text{ ft} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/wv49lz45tidm9a5laic5f91eixbq7opkqo.png)
Thus the maximum height reached by the ball is 70 ft