Question

A hot-air balloon begins rising from the ground at 4 meters per second at the same time a parachutist's chute opens at a height of 200 meters. The parachutist descends at 6 meters per second. Write a system of equations that represents the situation and solve it.

Answer 1

Since the rate of the rising hot air balloon is 4 meters per second, it will be at a certain distance
`4t` after t seconds. On the other hand, a descending parachutist travels a distance of
`200-6t` 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:


`\left \{ {{y=4t} \atop {y=200-6t}} \right.`

Then we can solve the system by equating both equations.


`4t=200-6t`

`10t=200`

`t=20`

ANSWER: The hot air balloon and the parachutist will be at the same height after 20 seconds.