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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.

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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.
User Babu James
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