Question

A balloon is rising at a constant speed of 5 ft/sec. A boy is cycling along a straight road at a constant speed of 15 ft/sec. When he passes under the balloon, it is 5 feet above him. Determine how fast the distance between the boy and the balloon is increasing 3 seconds after he has passed underneath it.

Answer 1

x = 15 t
h = 5 t + 5 = 5(t+1)

y^2=x^2+h^2 = 225t^2 + 25t^2 + 50 t + 25

y^2 = 250 t^2 + 50 t + 25

2 y dy/dt = 500 t + 50

now at t = 3
y^2 = 250(9)+150 +25 = 2425
y = 49.2
so
2(49.2) dy/dt = 1500 + 50 = 1550
dy/dt = 15.8 m/s

Answer 2

make a sketch putting the boy somewhere on the road

let the time passes since the boy was directly under the ballon be t seconds
the the boy went 15t ft and the balloon is at a height of (3 + 5t) ft
let the distance between them be d ft
so we have a right-angled triangle with base 15t, height of 3+5t and hypotenuse d

d^2 = (15t)^2 + (3+5t)^2
2d dd/dt + 2(15t)(15) + 2(3+5t)(5)
dd/dt = (225t + 15 +25t)/d
= (250t + 15)/d

when t=3
d^2 = 45^2 + 18^2 = 2349
d = √2349

dd/dt = (250(3) + 15)/√2349
= appr 15.78 ft/s

Answer 3

Go with Damon's solution, I somehow read the balloon as 3 ft instead of 5 ft above the boy

correction:
d^2 = (15t)^2 + (5+5t)^2
2d dd/dt + 2(15t)(15) + 2(5+5t)(5)
dd/dt = (225t + 25 +25t)/d
= (250t + 25)/d

when t=3
d^2 = 45^2 + 20^2 = 2425
d = √2425

dd/dt = (250(3) + 25)/√2425
= appr 15.738 ft/s