Homework Help: Hot-air balloon

Posted by Jake on Tuesday, March 15, 2011 at 3:15pm.

A hot-air balloon is 180 ft above the ground when a motorcycle passes directly beneath it (travelling in a straight line on a horizontal road) going at a constant speed of 60 ft/s. If the balloon is rising vertically at a rate of 15 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 20 seconds after the motorcycle was directly beneath the balloon.

This is what I understand.

d^2=h^2+x^2

(d)dd/dt= (h)dh/dt + (x)dx/dt

dd/dt= ((180)dh/dt + (60)dx/dt)/15

Other than that, I really have no idea what else to do. How do I find what dh/dt is as well as dx/dt.

Hot-air balloon - bobpursley, Tuesday, March 15, 2011 at 3:22pm
Did you read the jproblem? dh/dt=15ft/sec

dx/dt= 60 ft/sec

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