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A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angleof 30.0° from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air?

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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