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.
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Given u=18 km/hr = 5 m/s
time taken by the bag to reach ground can be calculated by using the formula
s=ut+(1/2)at²
10=(-5)t+1/2×9.81×t²
by solving this equation we can get 't' value
t=2.025 sec
force acing on the balloon, before bag is released
impulse momentum principle
(f-400×9.81)t=m(v-u)
here v=u
⇒ f=3924 N
final velocity of the balloon after the bag is dropped
here mass becomes 400-40=360 kg
this is because the bag was dropped and it weighs 40 kg
impulse momentum principle
(3924-360×9.81)t=m(v-u)
392.4×2.025=360(v-5)
⇒v=7.20725 m/s
time taken by the bag to reach ground can be calculated by using the formula
s=ut+(1/2)at²
10=(-5)t+1/2×9.81×t²
by solving this equation we can get 't' value
t=2.025 sec
force acing on the balloon, before bag is released
impulse momentum principle
(f-400×9.81)t=m(v-u)
here v=u
⇒ f=3924 N
final velocity of the balloon after the bag is dropped
here mass becomes 400-40=360 kg
this is because the bag was dropped and it weighs 40 kg
impulse momentum principle
(3924-360×9.81)t=m(v-u)
392.4×2.025=360(v-5)
⇒v=7.20725 m/s
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