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22 votes
22 votes
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?

User Brian Nezhad
by
2.8k points

2 Answers

16 votes
16 votes

Answer:Hector threw the ball 65 feet while Tre threw the ball 63 feet far

Explanation:

The farthest throw would probably be around 450 feet

User Sfk
by
3.3k points
19 votes
19 votes

Answer:


d=455.38~units

Explanation:

The coordinates of the dead center of the field,
D\equiv (322,322)

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

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


d=√((x_1-x_2)^2+(y_1-y_2)^2)


d=√((322-0)^2+(322-0)^2)


d=455.38~units

User Relet
by
2.7k points