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

Answer:

d=101 m

Explanation:

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

$t=\frac{h}{v}\\t=\frac{202m}{2.70m/s}\\\\t=\74.8s$

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.

$t=\sqrt{2\frac{202m}{9.8m/s^2}}\\t=6.4s$

the total time would be:

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

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