Question

A billiard ball is struck by a cue. It travels 100\,\text{cm}100cm100, start text, c, m, end text before ricocheting off a rail and traveling another 120\, \text{cm}120cm120, start text, c, m, end text into a corner pocket. The angle between the path as the ball approaches the rail and the path after it strikes the rail is 45^{\circ}45 ∘ 45, degrees. How far is the corner pocket from where the cue initially struck the ball?

Answer 1

about 86 cm

Explanation:

My understanding of the geometry is shown in the attached figure.

Since the angle of incidence is equal to the angle of reflection, the angle APP₂ will be 22.5°, and the vertical distance from the pocket P to the ball position B is ...

(PA +AB')sin(22.5°) = 220sin(22.5°) ≈ 84.19 . . . . cm

The horizontal distance PB is

(PA -AB)cos(22.5°) = 20cos(22.5°) ≈ 18.48 . . . . . cm

The distance PB is given by the distance formula ...

PB = √(84.19² +18.48²) ≈ 86.19 . . . . cm

The ball was initially about 86 cm from the corner pocket.

Answer 2

`\large \boxed{\text{86 cm}}`

Explanation:

I think what you are saying is represented by the diagram below.

The ball starts at A, rebounds from the far wall at a 45° angle and into the corner pocket at C.

You want the find the distance from A to C.

We can use the Law of Cosines to answer this question

`\begin{array}{rcl}b^(2) & = & a^(2) + c^(2) - 2ac\cos \theta\\& = & 120^(2) + 100^(2) - 2 *120*100 * \cos 45^(\circ)\\& = & 14400 + 10000 - 24000*0.7071\\& = & 24400 - 16971\\& = & 7429\\ b & = & \textbf{86 cm}\\\end{array}\\\text{The corner pocket is $\large \boxed{\textbf{86 cm}}$ from where the cue originally struck the ball.}`