A billiard ball is struck by a cue. It travels 100 cm before ricocheting off a rail and traveling another 120 cm 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 degrees.
How far is the corner pocket from where the cue initially struck the ball?
Do not round during your calculations. Round your final answer to the nearest centimeter.
+33653 With the angles of incidence and reflection both 45degrees, the ball travels along two sides of a right angled triangle. The distance between pocket and initial strike is therefore that of the hypotenuse of the triangle.
Distance = sqrt(100^2 + 120^2) cm.
+33653 Reading the question more carefully, I see I have misinterpreted the angles. The angle between incident and reflection is 45 degrees not 90 degrees, so we need to use the cosine rule:
a^2 = 100^2 + 120^2 - 2*100*120*cos(45°) where a is the distance between pocket and initial strike.
so a = sqrt(100^2 + 120^2 - 2*100*120*cos(45°))
a = 86 cm (to the nearest cm).
