Question

7. A high-speed digital camera captures a bungee jumper diving off a bridge. What happens to the distance between the jumper and the bridge in each picture taken?

Answer 1

in bungee jumping we know that the motion is like free fall

So let say that camera capture the pictures after every "t" interval of time

now here for the first "t" interval the free fall distance will be given as

$d_1 = \frac{1}{2} gt^2$

now for next interval the distance is

$d_2 = \frac{1}{2}g(2t)^2 - \frac{1}{2}gt^2 = \frac{3}{2}gt^2$

again next "t" seconds

$d_3 = \frac{1}{2}g(3t)^2 - \frac{1}{2}g(2t)^2= \frac{5}{2}gt^2$

so in each consecutive time interval the distance is increasing

and the distance of fall in each interval of time is given in the ratio of

1 : 3 : 5 : 7......................

so after each shot the distance will increase faster from the initial position