# 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 ?

##### 1 Answer

#### Answer:

The package reaches the ground in

#### 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 *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" = 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")#