Question

A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v= 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?

Answer 1

The wait time is
`t_w = 3.4723 \ s`

Step-by-step explanation:

From the question we are told that

The distance of the hot air balloon above the ground is
`z = 50 \ m`

The distance of the hot air balloon from the target is
`k = 100 \ m`

The speed of the wind is
`v = 15 \ m/s`

Generally the time it will take the balloon to hit the ground is

`t = \sqrt{ (2 * z )/(g) }`

where g is acceleration due to gravity with value
`g = 9.8 m/s^2`

substituting values

`t = \sqrt{ (2 * 50 )/(9.8) }`

`t = 3.194 \ s`

Now at the velocity the distance it will travel before it hit the ground is mathematically represented as

`d = v * t`

substituting values

`d = 15 * 3.194`

`d = 47.916 \ m`

Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as

`b = k - d`

substituting values

`b =100 -47.916`

`b = 52.084 \ m`

Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as

`t_w = (b)/(v)`

substituting values

`t_w = (52.084)/(15)`

`t_w = 3.4723 \ s`