Since the rate of the rising hot air balloon is 4 meters per second, it will be at a certain distance [tex]4t[/tex] after t seconds. On the other hand, a descending parachutist travels a distance of [tex]200-6t[/tex] where t is also in seconds. They will both be at a certain disrance y when they have traveled a common time t. Writing this in systems of equation we'll get:
[tex] \left \{ {{y=4t} \atop {y=200-6t}} \right. [/tex]
Then we can solve the system by equating both equations.
[tex]4t=200-6t[/tex]
[tex]10t=200[/tex]
[tex]t=20[/tex]
ANSWER: The hot air balloon and the parachutist will be at the same height after 20 seconds.
