Answer:
Four times traveling 40,12345 mts
Explanation:
We can express this as a sum as follows
[tex]\sum_{i=1}^4(10\cdot \frac{2}{3}^{i - 1} + 10\cdot \frac{2}{3}^{i } ) = 3250/81=40.12345679[/tex]
The first term is the dropping part and the second is the upward motion. this sums up to 40,12345 m after bouncing four times. As a program we could write using the while loop.
distance=0;
i:=0;
while distance<40 do
i:=i+1;
distance:=distance+ [tex]10\cdot \frac{2}{3}^{i - 1} + 10\cdot \frac{2}{3}^{i };[/tex]
end while;
Here the answer is in i and the closest distance traveled after reaching 40 is in the variable distance.
