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Consider that the earth is vacuum. Consider a person of weight 100 kg is falling from sky with an parachute . He is free falling at height of 3000 m. When will he reach the ground? What would happen if he opens the parachute in diving? Will it work ?

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Nope. :) Parachutes work because they provide a drag resistance, which doesn't exist in vacuum, because there isn't anything to provide the drag. – Kitchi Apr 12 '13 at 13:25
Didn't the Apollo astronauts drop a feather and a hammer on the moon to confirm that they arrive at the same time in vacuum. Similar concept here. – ja72 Sep 16 '13 at 13:03

You've assumed that the earth is in vacuum. By that phrase, I think you're mentioning that there's no atmosphere. Then, we can neglect the drag by which the guy attains the terminal velocity (which is about some 50 m/s). And, the mass 100 kg isn't necessary, since all objects fall at the same rate [1].

We all know the acceleration due to gravity is about $9.8\ \text{ms}^{-2}$. Though there's a slight variation of $g$ with altitude, we can approximate this to be 9.8 (or even 10 if you wanna have some guess on the time). This means that the velocity is 9.8 m/s after a second, and then 19.6 m/s and so on...

You've assumed the distance to be 3000 m. Using the basic equations of motion, it's very easy to calculate the time taken for the guy to crash into ground. The initial velocity being zero, $t=\sqrt{2s/10}\approx24\ s$...

[1]: I've neglected the use of parachute because both the parachute and the guy fall at the same rate (parachutes won't be useful in the absence of drag) which is already been testified by a historical experiment by Galileo (or the famous feather & coin)...

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A parachute wouldn't work in a vaccum, because of the absence of air. Parachutes work on the principal of prividing more drag, therefore decreacing speed. If there is no drag, a parachute is redundant. A better option would not to jump from 3000m heights in vaccums. ;)

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Can you explain what your answer adds to CrazyBuddy's answer? – Bernhard Apr 12 '13 at 15:57

NASA has a great website that discusses these issues, because they are also very important for space ships. Click here.

They didn't calculate $3000~\text{m}$ yet, but I think you might be able to do that yourself.

enter image description here

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