Question

In a billiards game, Pete hits a ball that is 20 in. from a wall. The ball travels 34 in. until it hits the wall and bounces to a position that is 16 in. from the wall. What is the distance, x, the ball traveled after it bounced off the wall to get to the ending position? Enter your answer, as a decimal, in the box.

Answer 1

Refer to the attached image,

Peter hits a ball that is 20 inch from a wall means AB=20 in

A ball travels 34 in until it hits the wall means AC=34 in

It bounces back to a position means CD = 'x' in (Say)

It bounces back to CD which is 16 in from the wall means DE=16 in

We have to calculate the value of 'x'.

Since, ball is reflected from point C. By law of reflection which states the angle of reflection is equal to the angle of incidence.

So, $\angle ACF=\angle FCD$ = y degrees (say)

Since angles BCF and FCE are right angles.

So, $\angle BCA=\angle DCE = 90-y^\circ$ (equation 1)

Now, in triangle ABC and CDE

$\angle B=\angle E$ (Each angle is right angle)

$\angle BCA=\angle DCE$ (From equation 1)

Therefore, triangle ABC is similar to triangle DEC by AA similarity.

So, the ratio of corresponding sides are equal.

$\frac{AB}{DE}=\frac{AC}{DC}$

Substituting the values,

$\frac{20}{16}=\frac{34}{x}$

$20x = 544$

x = 27.2 in

Therefore, the distance, x, the ball traveled after it bounced off the wall to get to the ending position is 27.2 in