Question

Newton tells us that gravitational force acts on all bodies in proportion to their masses. Why, then, doesn’t a heavy body fall faster than a light body?

help me solve @Abhisar

Answer 1

i believe it is because velocity is the same regardless of size.

Answer 2

u can ask two more. I promise !!

Answer 3

oh ok

Answer 4

V = U + at

Answer 5

u knw ths ?

Answer 6

????

Answer 7

no

Answer 8

okay then Velocity = accelaration X Time ?

Answer 9

yes

Answer 10

that's enough !!

Answer 11

Now on every body, doesn't matters if it is heavy or light, small or big, fat or thin, the accelaration acting due to gravity is 9.8 m/s^2

Answer 12

i see

Answer 13

So they will attain same velocity after a cetain time interval

Answer 14

Like suppose if a 10 kg body falls for 10 sec and 5 kg body also for 10 sec then final velocity for both will be 98m/s

Answer 15

got it

Answer 16

This is the reason why when a feather and a coin is dropped from a tower, they both reach the ground at same time

Answer 17

i see i see, you are awesome

Answer 18

So are u \(\color{green}{\huge\ddot\smile}\)

Answer 19

no more for today i am tired
maybe tomorrow. have a good rest of your day

Answer 20

Thanx and Same to uh !

Answer 21

The feather might not fall as fast because of greater air resistance! But don't let it trick you :)

Here's another explanation, which tells you why acceleration will be the same for both.

Do you know of Newton's Universal Gravitation Law? It describes the force between two masses due to gravitation.
It is
\(F=G\dfrac{m\ M}{r^2}\)
where
\(m\) is one mass, \(M\) is the other. We'll say \(M\) is th Earth, so \(m\) is an object falling.
\(G\) is just a constant (very, very small)
and \(r\) is the distance between the masses' centers of gravity (I think)

On another note Newton also said something like \(F_\text{net}=m\ a\), which I'm sure you know. So, the acceleration component due to gravity is often written as \(g\), and so we say that
\(F_\text{gravity}=m\ g\)

Now
\(F_\text{gravity}=m\ g=G\dfrac{m\ M}{r^2}\)

So, these two laws allow us to see something neat. There are\(m\)'s in the middle and right expressions above. So we cancel them by division. Then (and I'll right all terms to keep track of them)
\(\dfrac{F_\text{gravity}}m=g=G\dfrac{M}{r^2}\)

The left and middle expressions are consistent with our \(F_\text{gravity}=m\ g\), so that's good. However, the equation on the right is rather nifty!
\(\large\color{green}{g=G\dfrac{M}{r^2}}\)
This shows that \(\color{green}g\) doesn't depend on \(m\), the mass of the object. So, we can say that the acceleration due to gravity is the same for the same distance between centers of gravity and same other mass. And this is the acceleration difference between, say, the Earth and a falling object.

Answer 22

By the way, the gravitational acceleration is about 9.8 meters per second squared at sea level and the area a little above or below sea level. It gets less, though, as you go into space. Astronauts in space do not experience 9.8m/s^2!