Hotair balloon
posted by Jake .
A hotair 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.

Hotair balloon 
bobpursley
Did you read the jproblem? dh/dt=15ft/sec
dx/dt= 60 ft/sec
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