Question

A hot-air balloon is ascending at the rate of 12 m/s and is 51 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground ?

Answer 1

The package reaches the ground in `"4.67 s"` and hits the ground with a speed of `"33.8 m/s"`.

Explanation:

The idea here is that once it is released from the balloon, the package will aquire the velocity of the balloon.

This means that the package will ascend with an initial velocity of `"12 m/s"` oriented upward until it comes to a complete stop at maximum height, then starts free falling towards the ground.

The total time needed for the package to reach the ground will have to include the time it takes the package to ascend to maximum height and the time it takes the package to fall to the ground.

So, at maximum height the package has a velocity equal to zero, which means that you can write

`underbrace(v^2)_(color(blue)(=0)) = v_0^2 - 2 * g * h_"up"`

Rearrange to solve for `h_"up"`

`h_"up" = v_o^2/(2 * g) = (12^2 "m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = "7.35 m"`

The time it takes for the package to ascend 7.35 m is

`underbrace(v)_(color(blue)(=0)) = v_0 - g * t_"up"`

`t_"up" = v_0/g = (12color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = "1.22 s"`

The package reaches a maximum height of

`h_"max" = h + h_"up"`

`h_"max" = 51 + 7.35 = "58.35 m"`

The time it takes the package to free fall from this height is

`h_"max" = underbrace(v_"top")_(color(blue)(=0)) * t_"down" + 1/2 * g * t_"down"^2`

`t_"down" = sqrt( (2 * h_"max")/g) = sqrt((2 * 58.35color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^2)) = "3.45 s"`

The total time it takes the package to reach ground, from the moment it's released from the balloon, is

`t_"total" = t_"up" + t_"down" = 1.22 + 3.45 = color(green)("4.67 s")`

The speed with which it reaches the ground is

`v_"bottom" = underbrace(v_"top")_(color(blue)(=0)) + g * t_"down"`

`v_"bottom" = 9.8"m"/"s"^color(red)(cancel(color(black)(2))) * 3.45color(red)(cancel(color(black)("s"))) = color(green)("33.8 m/s")`