Given that if a ball is dropped from x feet, it bounces up to 2/3 x feet.
And the ball is dropped from 10 feet, that is x=10 feet,
So,before the first bounce it travels 10 feet distance.
Between first and second bounce it travels 
Between second and third bounce, it travels 
Between third and fourth bounce, it travels 
Like that between 29th and 30th bounce, it travels 
Hence total distance traveled is

=![10+20[(\frac{2}{3}) +(\frac{2}{3}) ^{2} +(\frac{2}{3}) ^{3} +.....+(\frac{2}{3} )^{29} ]](https://tex.z-dn.net/?f=10%2B20%5B%28%5Cfrac%7B2%7D%7B3%7D%29%20%2B%28%5Cfrac%7B2%7D%7B3%7D%29%20%5E%7B2%7D%20%2B%28%5Cfrac%7B2%7D%7B3%7D%29%20%5E%7B3%7D%20%2B.....%2B%28%5Cfrac%7B2%7D%7B3%7D%20%29%5E%7B29%7D%20%5D)
= ![10+20[\frac{\frac{2}{3}*(1-(\frac{2}{3})^{29}) }{1-\frac{2}{3} }]](https://tex.z-dn.net/?f=10%2B20%5B%5Cfrac%7B%5Cfrac%7B2%7D%7B3%7D%2A%281-%28%5Cfrac%7B2%7D%7B3%7D%29%5E%7B29%7D%29%20%20%7D%7B1-%5Cfrac%7B2%7D%7B3%7D%20%7D%5D)
= 10+20*2*(1-
)
= 49.9997 feet ≈ 50 feet approximately.