# Dragged fall before reaching terminal velocity

Consider two objects A and B (of the same shape and size) in fall with air resistance. Object A is much heavier than B. I learned that since object A has a higher terminal velocity it reaches the ground before B. I was wondering if both objects hit the ground before they reach the terminal velocity of B, then they would both hit the ground at the same time. Correct me if i'm wrong. Sorry if it was a basic question, just eager to learn.

• You say objects are of same shape and material but of different weights. This means one must be bigger than the other, right? – Deep Nov 25 '17 at 10:41
• I have edited the question. – Sanstroke Nov 25 '17 at 11:09

Well it is common knowledge that objects of the same shape will fall down to earth with the same acceleration regardless of their mass. This can be seen from Newton's equation for the force of gravity: $$F_{grav}=\frac{GMm}{r^2}$$ where $G$ is the gravitational constant, $M$ is the mass of earth and $m$ is the mass of the object. As $F=ma$, we can see that $a=\frac{F}{m}$, and so the acceleration of the objects will be: $$a=\frac{GM}{r^2}$$We can see from this that the objects themselves have nothing to do with the acceleration due to gravity.

Now terminal velocity comes about due to that fact that when objects fall through a fluid, such as air, they experience a drag force on them. This drag force is a function of velocity, so as the velocity increases, so does the drag force. Terminal velocity is reached when the drag force on the object acting upwards is equal to the gravitational force pulling it downwards, and so the net force is $0$. Newton's second law states that:

"An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force"

As the forces here are balanced we can see that once the object has reached it's terminal velocity it will no longer increase in speed.

This means that if $B$ has a lower terminal velocity than $A$, it will stop accelerating before $A$, and $A$ will continue until it reaches it's higher terminal velocity. From this we can see that $A$ will hit the ground before $B$, as long as $B$ reaches its terminal velocity.

From this we could say that if $B$ hasn't reached it's terminal velocity, they should both hit the ground at the same time, but we should think about why terminal velocity comes about - due to the drag force. If $B$ has a lower terminal velocity than $A$, then it's clear that $B$ experiences more drag than $A$, and therefore it will not accelerate towards the ground at the same speed as $A$, because the force of drag pushing it up is greater, so the net force downwards is smaller.

If we go with this then we can say that as long as $A$ and $B$ are in the same fluid, $A$ will hit the bottom before $B$.

• Thanks for replying. But I find "If B has a lower terminal velocity than A, then it's clear that B experiences more drag than A" difficult to believe. I think B has lower terminal Velocity because it is lighter( the downward force is lower so lesser drag balances the downward force, creating a terminal velocity). – Sanstroke Nov 25 '17 at 11:07
• That is very true - sorry I just assumed that they had the same mass, didn't read the question properly (Doh!). – CooperCape Nov 25 '17 at 11:15
• @Sanstroke If the two objects are the same shape then as long as $B$'s terminally velocity isn't reached they will hit the ground at the same time, otherwise $A$ should hit the ground first. – CooperCape Nov 25 '17 at 11:30