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.
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?
1 answer