Answer:
d=101 m
Explanation:
The time taken to reach the 202m is given by:
[tex]t=\frac{h}{v}\\t=\frac{202m}{2.70m/s}\\\\t=\74.8s[/tex]
the problem states that the balloonist maintain that altitude for 10.3 seconds more, so:
[tex]t=74.8s+10.3s=85.1s[/tex]
if the bag falls straight down:
[tex]d=v*t\\d=1.1m/s*85.1s\\d=93.6m[/tex]
if the bag is affected by the velocity of the wind we need to calculate the time that the bag takes to reach the ground.
[tex]t=\sqrt{2\frac{202m}{9.8m/s^2}}\\t=6.4s[/tex]
the total time would be:
[tex]t_t=85.1+6.4\\t_t=91.5s[/tex]
[tex]d=v*t_t\\d=1.1m/s*91.5s\\d=101m[/tex]
