Question

A ball is released from a 10 m high roof and bounces two thirds as high on each successive bounce. After traveling a total of 40 m (up and down motion), how many times did the ball bounce (e.g. touch the ground)? Put the answer in

Answer 1

Answer:

Four times traveling 40,12345 mts

Explanation:

We can express this as a sum as follows

$\sum_{i=1}^4(10\cdot \frac{2}{3}^{i - 1} + 10\cdot \frac{2}{3}^{i } ) = 3250/81=40.12345679$

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+ $10\cdot \frac{2}{3}^{i - 1} + 10\cdot \frac{2}{3}^{i };$

end while;

Here the answer is in i and the closest distance traveled after reaching 40 is in the variable distance.