Question

A hot air is attached to a spool of rope that is 125 feet away from the balloon when it is on the ground. The hot air balloon rises straight up in such a way that the length of rope increases at a rate of 15 ft/sec. How fast is the hot air balloon rising 20 seconds after it lifts off ?

Answer 1

`16.5\text{ ft per sec}`

Explanation:

Let the length of rope is x and y be the distance of the balloon from the ground,

Given,

The spool of rope that is 125 feet away from the balloon,

By making the diagram of this scenario,

Using Pythagoras theorem,

`x^2 = y^2 + 125`

Differentiating with respect to time (t),

`2x(dx)/(dt)=2y(dy)/(dt)+0`

We have,
`(dx)/(dt)=15\text{ ft per sec}`

If time = 20 seconds,

So, x = 15 × 20 = 300 ft ( ∵ distance = speed × time )

By substituting the values,

`2* 300* 15 = 2y(dy)/(dt)`

`\implies (dy)/(dt)=(4500)/(y)`

Since,
`x^2 = y^2 + 125^2` and x = 300 ft,

`300^2 = y^2 + 125^2`

`90000 - 15625 = y^2`

`74375 = y^2`

`\implies y = √(74375)`

`\implies (dy)/(dt)=(4500)/(√(74375))\approx 16.5\text{ ft per sec}`

Hence, the hot air balloon is rising with the speed of 16.5 ft per sec.