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Sagot :
Given data:
* The initial velcoity of the ball is 40 m/s.
* The angle made by the velocity with the horizontal is 30 degree.
Solution:
The horizontal range of the projectile in terms of the initial velocity is,
[tex]R=\frac{u^2\sin (2\theta)}{g}[/tex]where R is the horizontal range, u is the initial velocity, g is the acceleration due to gravity, and
[tex]\theta\text{ is the angle made by velocity with horizontal}[/tex]Substituting the known values,
[tex]\begin{gathered} R=\frac{40^2\times\sin(2\times30^{\circ})^{}}{9.8} \\ R=141.4\text{ m} \end{gathered}[/tex]Thus, the ball travel 141.4 m in the horizontal range.
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