Suppose you drop a ball from a window 30 meters above the ground. The ball bounces to 50% of its previous height with each bounce. What is the total number of meters (rounded to the nearest tenth) the ball travels between the time it is dropped and the ninth bounce?
+37091 Before it bounces it travels 30 m
betweent the first and second bounce it goe UP 15 m and down 15m then the second bounce occurs.....etc
We are not counting the distance it travels AFTER the 9th bounce (per the question)
30+ (2)(30 x .5) + (2) 30 x .5 x .5 +(2) 30 x .5x.5x.5 .....= 30+ 60 (.5 + .5x.5 + .5x.5x.5....) =30+ 60\(\sum_{1}^{8}\) .5^n =
30 +60(.99609) = 89.765 m
Can anyone verify this????
Thanx
+37091 Before it bounces it travels 30 m
betweent the first and second bounce it goe UP 15 m and down 15m then the second bounce occurs.....etc
We are not counting the distance it travels AFTER the 9th bounce (per the question)
30+ (2)(30 x .5) + (2) 30 x .5 x .5 +(2) 30 x .5x.5x.5 .....= 30+ 60 (.5 + .5x.5 + .5x.5x.5....) =30+ 60\(\sum_{1}^{8}\) .5^n =
30 +60(.99609) = 89.765 m
Can anyone verify this????
Thanx
