Physics: Comparing Stone Drops from 100m Building

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When two stones of different masses are dropped from a 100 m building, they will fall the same distance in a vacuum due to gravity, which accelerates all objects equally regardless of mass. The distance fallen can be calculated using the equation D = 0.5 * g * t^2, resulting in approximately 122.625 m after 5 seconds, assuming no air resistance. In real-world conditions, air friction can cause slight differences in distance traveled, with heavier stones potentially experiencing less relative deceleration. The discussion emphasizes that while mass does not affect gravitational acceleration, factors like shape and density influence the effects of air resistance. Ultimately, understanding these principles requires isolating variables to grasp their individual impacts on motion.
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I have one doubt in physics...
If two stones(one stone is having 10 kg mass and other one is having 5 kg)are drop down from the 100 m building. After 5 second, how much distance they will cover ?
 
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The mass of the stones does not affect the rate at which they fall. This was one of the idealogical breakthroughs of Newtonian mechanics. The acceleration any object feels from the gravitational pull of the Earth is only dependent on the mass of the Earth, and not on the mass of the body being pulled - meaning everything gets pulled at the same rate.

Because of this, the two stones would fall the exact same distance in 5 seconds. The distance is determined by the equation ##(Distance \ Fell) = \frac{1}{2} \cdot (Acceleration \ Due \ to \ Earth) \cdot (Time \ Elapsed)^2##.

We use ##g## to represent the acceleration due to Earth at the surface of the Earth so this equation simplifies to ##D = \frac{1}{2} \cdot g \cdot t^2##.
##g## is equal to about 10 (in the units we are using here), so the distance traveled is roughly ##\frac{1}{2} \cdot 10 \cdot 25 = 125 \ meters##.

The important thing to take away is that the acceleration a body feels in a gravitational feel is only dependent on the mass of the thing pulling the body, not on the mass of the thing being pulled.
 
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Hardik Batra said:
I have one doubt in physics...
If two stones(one stone is having 10 kg mass and other one is having 5 kg)are drop down from the 100 m building. After 5 second, how much distance they will cover ?

If the air friction is ignored both of them will have the same acceleration. If the building is on the Earth the acceleration is approximately 9.81 m/(s^2).

If the initial velocities are zero the distance can be calculated by the equation:

x = (1/2).a.t^2

x = 0.5 x 9.81 x 5^2

x = 122.625 m

However as the height of the building is 100 m both of them will travel 100 m. At the end both of them will be on the ground.

If the air friction is not ignored, the shape of the stones and the density of the air are required to calculate the distance.
 
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Your mean to say both will cover same distance..

But By practically 10 kg stone will cover more distance than 5 kg.

What to understand in this?
 


Practically, there is a small difference due to air friction as Zalajbeg said. The 'slowing' force which air imparts onto a falling object does depend on the characteristics of the object itself - including it's mass (density would be a better term).

But, in a vacuum, there is no air resistance, and therefore two objects of different weights (even drastically different weights) will fall the exact same distance.

A lovely example of this is the following clip where a feather and a hammer are dropped in the airless environment of the moon and they fall at the exact same rate.
 
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Hardik Batra said:
Your mean to say both will cover same distance..

But By practically 10 kg stone will cover more distance than 5 kg.

What to understand in this?

One stone may not have the same density as another stone. This is another factor if air resistance is to be taken into account - because the volumes of the stones or their cross sectional areas are what count.
The essence of getting to grips with Physics is to consider, whenever possible, the various effects one at a time. There are many examples where it becomes impossible to discuss some of the simplest phenomena when you add more and more extra variables into the model. Once each effect has been understood on its own then it's possible to combine them into a 'real' practical situation.
 


Here is a fun and classic piece of experimental evidence;

 
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