Question

A hot-air balloon is ascending at the rate of $12\dfrac{m}{s}$ and is $51m$ 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

Firstly, we will apply the third equation of motion to find the maximum height, by putting the value of velocity equal to zero. Then with the help of the first equation of motion, we will get the value of the time required for the package to ascend.

Complete step by step answer:
The concept which is being used here is that once it is released from the balloon, the package will move with the velocity of the balloon.
This implies that the package will ascend with an initial velocity of $12\dfrac{m}{s}$ oriented upward until it stops at maximum height and then starts free falling towards the ground.
So, at maximum height the package has a velocity which is equal to zero, which means that we can write that,
${v^2} = v_o^2 - 2g{h_{up}}$
As $v = 0\dfrac{m}{s}$,
${h_{up}} = \dfrac{{v_o^2}}{{2g}}$
On putting the required values,
${h_{up}} = \dfrac{{{{12}^2}}}{{2 \times 9.8}}$
On solving this, we get,
${h_{up}} = 7.35m$
Now, we will calculate the time in which the package ascends,
$v = {v_o} - g{t_{up}}$
Again we know that $v = 0\dfrac{m}{s}$
${t_{up}} = \dfrac{{{v_o}}}{g}$
${t_{up}} = \dfrac{{12}}{{9.8}}$
${t_{up}} = 1.22s$
Now, the maximum height the package reaches is,
${h_{\max }} = h + {h_{up}}$
${h_{\max }} = 51 + 7.35$
${h_{\max }} = 58.35m$
The time taken by the package to free fall from this height is,
${h_{\max }} = {v_{top}}.{t_{down}} + \dfrac{1}{2}gt_{down}^2$
On putting the required value, we get,
${t_{down}} = \sqrt {\dfrac{{2.{h_{\max }}}}{g}}$
${t_{down}} = \sqrt {\dfrac{{2 \times 58.35}}{{9.8}}}$
${t_{down}} = 3.45\sec$
So, the total time taken by the package to reach the ground is,
${t_{total}} = {t_{up}} + {t_{down}}$
On putting the required values,
${t_{total}} = 1.22 + 3.45$
${t_{total}} = 4.67\sec$
The speed at which the package reaches the ground is,
${v_{bottom}} = {v_{top}} + g.{t_{down}}$
We know that the velocity of the package at the top point is equal to zero, so,
${v_{bottom}} = g.{t_{down}}$
${v_{bottom}} = 9.8 \times 3.45$
${v_{bottom}} = 33.8\dfrac{m}{s}$
So, the time in which the package reaches the ground is ${t_{total}} = 4.67\sec$ and the speed which it hits the ground is ${v_{bottom}} = 33.8\dfrac{m}{s}$.

Note: The total time which is needed for the package to reach the ground will include the time it takes the package to ascend to maximum height and then the time it takes for the package to fall to the ground.