## Question:

A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m/s . If the bag is released with the same upward velocity of 6 m/s when t = 0 and hits the ground when t = 8 s , determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant.

## Step-by-step

\begin{aligned}(+ \downarrow)\quad\quad & s = { s }_{ 0 } + { v }_{ 0 }t + \frac { 1 } { 2 } { a }_{ c } { t }^{ 2 } \\ & h = 0 + (-6)(8) + \frac { 1 } { 2 }(9.81){ (8) }^{ 2 } \\ & \space\space = 265.92 \text{ m } \end{aligned}

During t = 8 s , the balloon rises

$h’ = vt = 6(8) = 48 \text{ m }$

Altitude = $h + h’ = 265.92 + 48 = 314 \text{ m }$

\begin{aligned}(+ \downarrow)\quad\quad & v = { v }_{ 0 } + { a }_{ c }t \\ & v = -6 + 9.81(8) = 72.5 \text{ m/s } \end{aligned}