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A small boy is throwing a ball towards a wall 6 in front of him. He releases the ball at a height of $1.4 \mathrm{~m}$ from the ground. The ball bounces from the wall at a height of $3 \mathrm{~m}$, rebounds from the ground and reaches the boy's hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground?
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The correct answer is:
$1.5 \mathrm{~m}$
$3=(6+x) \tan \theta\left[1-\frac{6+x}{12}\right]$
$3=\frac{(6+x)(6-x) \tan \theta}{12}$
$1.4=x \tan \theta\left[1-\frac{x}{12}\right]$
$1.4=\frac{x(12-x)}{12} \tan \theta$
$\frac{30}{14}=\frac{36-x^{2}}{12 x-x^{2}}$
$360 x-30 x^{2}=36 \times 14-14 x^{2}$
$16 x^{2}-360 x+36 \times 14=0$
$x=\frac{360 \pm \sqrt{(360)^{2}-4 \times 36 \times 14 \times 16}}{32}$
$x=\frac{360 \pm 312}{32}=\frac{48}{32}=1.5$
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