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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?

User Tobixen
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1 Answer

2 votes

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,


R=(u^2\sin (2\theta))/(g)

where R is the horizontal range, u is the initial velocity, g is the acceleration due to gravity, and


\theta\text{ is the angle made by velocity with horizontal}

Substituting the known values,


\begin{gathered} R=\frac{40^2*\sin(2*30^(\circ))^{}}{9.8} \\ R=141.4\text{ m} \end{gathered}

Thus, the ball travel 141.4 m in the horizontal range.

User Amlxv
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