A squash ball player hits the ball 2.3m to the side wall. The ball rebounds at an angle of 100 and travels 3.1m to the front wall.
How far is the ball from the player when it hits the front wall? ( Assume the player does not move after the shot.)
1 Answer | Add Yours
The squash player hits the ball 2.3 m to the side wall. The ball hits the side wall and rebounds at an angle of 100 degrees. It then travels 3.1 m to the front wall.
When the ball hits the side wall and rebounds, the angle made with the normal on striking is 50 degrees. This gives the distance moved towards the front wall as X, where tan 50 = X/2.3
=> X = 2.3*tan 50
=> X = 2.741 m
There is right triangle made by the ball in the path it takes as it moves after striking the side wall. The hypotenuse is 3.1 m. If the distance by which it moves away from the player is Y, sin 50 = Y/3.1
=> Y = 3.1* sin 50
=> Y = 2.374 m
X + Y = 5.115 m
The distance of the ball from the player when it strikes the front wall is 5.115 m
We’ve answered 318,955 questions. We can answer yours, too.Ask a question