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Tre and Hector want to calculate the maximum possible throw at this field.


They calculate that the farthest point on the field would be the center fielder standing back at the dead center wall at point (322, 322). Suppose the center fielder threw the ball from here to home base.


How far is the throw?

Respuesta :

Answer:

[tex]d=455.38~units[/tex]

Step-by-step explanation:

The coordinates of the dead center of the field, [tex]D\equiv (322,322)[/tex]

Home base in a coordinate system is always the origin, [tex]O\equiv(0,0)[/tex]

The center fielder threw the ball form the dead center to the home base, hence the distance of throw:

[tex]d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]

[tex]d=\sqrt{(322-0)^2+(322-0)^2}[/tex]

[tex]d=455.38~units[/tex]

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