We appreciate your visit to On a baseball field the pitcher s mound is 60 5 feet from home plate During practice a batter hits a ball 186 feet at. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Answer:
The outfielder threw the ball 145 feet away.
Step-by-step explanation:
We are given that,
Distance of mound from home plate = 60.5 feet
Distance of outfielder from home plate = 186 feet
Angle made by the batter = 40°
So, using the cosine law, we have,
[tex]c^{2}=a^{2}+b^{2}-2ab\cos \theta[/tex]
i.e. [tex]c^{2}=(60.5)^{2}+(186)^{2}-2\times 60.5\times 186\times \cos 40[/tex]
i.e. [tex]c^{2}=38256.25
-22506\times 0.766[/tex]
i.e. [tex]c^{2}=38256.25
-17239.596
[/tex]
i.e. [tex]c^{2}=21016.654
[/tex]
i.e. c = ±144.97 ≈ ±145
Since, the distance cannot be negative.
Thus, the outfielder threw the ball 145 feet away.
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Rewritten by : Barada
Use cosine law to solve for x
[tex]x^2=60.5^2+186^2-2\times60.5\times186\times \cos{40^o}
\\x^2=38,256.25-17,240.6
\\x^2=21,015.65
\\x= \sqrt{21,015.65}
\\x=145[/tex]
