# A balloon is 20m horizontally away from an observer. At an instant, the balloon rises vertically above the ground, the distance between the observer and the balloon changes at 1.25 m/sec.

How fast is the balloon rising vertically?

## At a time of t seconds,

let the vertical height be h m and

let the distance between the observer and the balloon be x

x^2 = h^2 + 20^2

2x dx/dt = 2h dh/dt + 0

x(1.25) = h dh/dt

dh/dt = 1.25x/h

We have to know either h or x at that instant, then use my first

equation to find the second variable.