The horizontal range of the ball is 6.13 m
Explanation:The initial velocity of the ball, u = 8 m/s
Angle made by the ball with the horizontal, θ = 35°
Acceleration due to gravity, g = 9.81 m/s²
The horizontal range is calculated as shown below
[tex]R=\frac{u^2\sin 2\theta}{g}[/tex]Substitute u = 8 m/s, θ = 35°, and g = 9.81 m/s² into the range formula above
[tex]\begin{gathered} R=\frac{8^2\sin (2\times35)}{9.81} \\ R=\frac{64\sin (70)}{9.81} \\ R=\frac{64(0.9397)}{9.81} \\ R=6.13\text{ m} \end{gathered}[/tex]Therefore, the horizontal range of the ball is 6.13 m
