Answer to Question #147346 in Calculus for Sean

Question

A small balloon is released at a point 150 ft away from an observer who is on level ground.
If the balloon goes straight up at the rate of 8 ft/sec, how fast is the distance from the observer to the balloon increasing when the balloon is 50 ft high?

Answer 1

Initial distance between observer and balloon is d₀=150 ft;

the speed of the balloon is v=8 ft/sec;

when balloon is h=50 ft high, final distance can be calculated as;

d_f=(h²+d²)^1/2=(150²+50²)^1/2=158.114 ft;

time to reach 50 ft height is t= h/v=50/8=6.25 sec;

the speed of the balloon with respect to the observer is V_bo:

V_bo=(d_f-d₀)/t=8.114/6.25=1.298 ft/sec;

Answer:

the speed of the balloon with respect to the observer is 1.298ft/sec