Question

The balloon has a total mass of 400 kg including the passengers and ballast. The balloon is rising at a constant velocity of 18 km/hr when h = 10 m. If the man drops the 40-kg sand bag, determine the velocity of the balloon when the bag strikes the ground. Neglect air resistance.

Answer 1

balloon is rising at a constant velocity of 18 km/h
If man drop the 40kg sand bag , then initial velocity of sand bag , u = 18 km/h
e.g., u = 18 × 5/18 = 5 m/s

Let t is the time taken by sand bag to strike the ground.
use formula,
S = ut + 1/2at²
Here, h = -10 m [ ∵ body moves downward the reference level h m below ]
a = - g = -10 m/s² and u = 5 m/s²
∴ -10 = 5t - 1/2 × 10 × t²
⇒ -2 = t - t²
⇒ t² - t - 2 = 0
t = 2 ∵ t ≠ -1

Hence, time taken by sand bag to strike the ground is 2 sec

Now, use impulse and momentum concept.
After removing 40kg sand bag from 400 kg of balloon , impulsive force acts and rest mass of Balloon change their momentum .
so, impulse = change in momentum
∴ F.t = m(v - u)
Here , F is the net force act on sand bag
e.g., F = mass of sand bag × g = 40 × 10 = 400
m = rest mass on balloon after removing sand bag .
m = 400 - 40 = 360 kg

∴ 400 × 2 = 360 (v - 5)
⇒80 = 36(v - 5)
⇒ 20 = 9(v - 5)
⇒2.2 = v - 5
v = 7.2 m/s

Hence, velocity of sand bag when the bag strikes the ground = 7.2 m/s

Answer 2

Mass of balloon = 400kg. (includes everything)
Initial velocity = 18km/hr = 18 * 5/18 m/s = 5m/s.

Consider sign convention as upwards and leftwards to be negative and downwards and rightwards to be positive.

So, initial velocity (u) = +5 m/s
Also, acceleration due to gravity (g) = -g = -10 m/s
[Note: Only acceleration in this question is the acceleration due to gravity]
We need to find the time. Let the change in time be t.
Given, height till which the sandbag needs to drop = -10m

Applying the formula,
$S = ut + \frac{1}{2}g t^{2}$

We get,
$-10 = 5t - \frac{1}{2} * 10 * t^{2}$
⇒$-2=t- t^{2}$
⇒$(t+1)(t-2) = 0$
As we can't go reverse time, accepted time change = 2s.

So the sandbag takes 2s time to reach the ground.

We need to find the velocity at this interval for the balloon.
Applying impulse-momentum concept,

We know,
FΔt = Δ(mv)

Force applied = sandbag mass * g = 40*10 = 400N
Rest mass of balloon after dropping the sandbag = 400-40 = 360kg
⇒$400 * 2 = 360 * (v - 5)$
⇒$v-5 = \frac{400*2}{360} = \frac{20}{9} = 2.22$
⇒$v=2.2+5 = 7.2$

Hence, the velocity of the balloon when sandbag hit the ground = 7.2m/s

[ANSWERED]