Question

Experiencing a constant horizontal 1.10 m/s wind, a hot-air balloon ascends from the launch site at a constant vertical speed of 2.70 m/s. At a height of 202 m, the balloonist maintains constant altitude for 10.3 s before releasing a small sandbag. How far from the launch site does the sandbag land?

Answer 1

d=101 m

Step-by-step explanation:

The time taken to reach the 202m is given by:

`t=(h)/(v)\\t=(202m)/(2.70m/s)\\\\t=\74.8s`

the problem states that the balloonist maintain that altitude for 10.3 seconds more, so:

`t=74.8s+10.3s=85.1s`

if the bag falls straight down:

`d=v*t\\d=1.1m/s*85.1s\\d=93.6m`

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.

`t=\sqrt{2(202m)/(9.8m/s^2)}\\t=6.4s`

the total time would be:

`t_t=85.1+6.4\\t_t=91.5s`

`d=v*t_t\\d=1.1m/s*91.5s\\d=101m`