Question 53765
A hot air balloon is headed due east at a constant altitude of 200 feet, going 30 mph steady speed. It passes over Rover the dog, who is sitting still. Find an expression for the distance (in feet) from Rover to the balloon as a function of time (in seconds).
I figured out that 30 mph= 44.02 feet per second.
Also, I know to use the pythag thereom, but I can't simplify it from there. Help!
LET ROVER BE AT R ON GROUND.
LET V BE THE VERTICAL POSITION OF BALLOON FROM ROVER.
LET THE BALLON BE AT P AFTER T SECS.
TRIANGLE VRP IS RIGHT ANGLED AT R.HENCE 
RP^2 = RV^2+VP^2
RV= 200'
VP = TIME * VELOCITY = T*30 MPH =T*30*22/15 = 44T FT
RP^2 = 200^2+(44T)^2=40000+1936T^2
RP = [1936T^2+4000}60.5