A hard ball dropped from a height of 1 m in earth’s gravitational field bounces to a height of 95 cm. What
will be the total distance traversed by the ball?
Counting all bounces? That would be
100 (1 + 2*0.95 + 2*0.95^2 + 2*0.0.95^3 + ... centimeters. That is an infinite series, and assumes that the down-to-up ratio is always 0.95.
That can be written as
100*(2 + 2*0.95 + 2*0.95^2 + 2*0.0.95^3....) - 100
or, in closed form, as
200[1/(1-0.95)] - 100
= 3900 cm
A question like this was answered recently. I'll try to find a link.
drwls is faster than me!!
To calculate the total distance traversed by the ball, we need to take into consideration the initial drop and subsequent bounce. The ball is dropped from a height of 1 m and bounces back to a height of 95 cm.
First, let's calculate the distance covered during the drop. The ball falls from a height of 1 m, so the distance covered during the drop is 1 m.
Next, let's calculate the distance covered during the bounce. The ball reaches a maximum height of 95 cm, which is equivalent to 0.95 m. Since the ball bounces back to this height, it covers the same distance on the way down and again on the way up. Therefore, the total distance covered during the bounce is 2 times the height, which is 2 * 0.95 m = 1.9 m.
Finally, let's sum up the distance covered during the drop and the bounce:
Distance covered during the drop = 1 m
Distance covered during the bounce = 1.9 m
Total distance traversed by the ball = Distance covered during the drop + Distance covered during the bounce
Total distance traversed by the ball = 1 m + 1.9 m = 2.9 m
Therefore, the total distance traversed by the ball is 2.9 meters.