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A narrow arch supporting a stone bridge is in the shape of half of an ellipse and is 24 meters long and 8 meters high. A person standing at one location throws a rubber ball against the arch. No matter what direction the ball is thrown it always bounces off the arch once and strikes the same point on the ground. What is the distance between he person throwing the ball and the point on the ground at which the ball first strikes

Answer :

okpalawalter8

Answer:

[tex]X=17.9m[/tex]

Step-by-step explanation:

Length of arc [tex]L_a=24m[/tex]

Height of arc [tex]H_a=8m[/tex]

Tet the length on the ground be

[tex]2a=24\\ a=12m[/tex]

Generally the Pythagoras equation is mathematically given by

[tex]a^2=b^2+c^2[/tex]

[tex]c^2=a^2+b^2[/tex]

[tex]c^2=12^2+8^2[/tex]

[tex]c^2=144^2+64^2[/tex]

[tex]c=8.9[/tex]

Therefore the distance between he person throwing the ball and the point on the ground at which the ball first strikes X

[tex]X=2c\\X=2*8.9[/tex]

[tex]X=17.9m[/tex]

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