# Air resistance: how does it impact the final velocity?

In an ideal situation where air resistance is not considered, if you project an object upwards vertically with a certain velocity, it falls back down with the same velocity, but in the opposite direction.

$V2 = u2 + 2as$ proves this as the displacement becomes zero. $V=u$ or $V=-u$, the latter is the suitable answer because the velocity is in the opposite direction

Now in the same scenario, if we consider air resistance, my teacher tells me that the final velocity does not equal that of the first. (Magnitude wise) But doesn't the equation of motion stated above contradict this?

The presence of air resistance will only impact acceleration, but as it comes to the starting point, when it falls down, hence resulting in a zero displacement. Thus v= -u according to the above equation.

Have I missed something? Please excuse me if there is some misconception

• You derived that equation only in the absence of friction... – ACuriousMind Aug 30 '16 at 19:07

But doesn't the equation of motion stated above contradict this?

Yes, but that equation is not suitable for a case with air resistance. Air resistance will make the acceleration depend on the velocity rather than be constant.

Since the velocity of the ball is different on the way up and the way down, the acceleration curves are different on the way up and the way down. In particular on the way up, the downward acceleration will be greater than $g$, while on the way down the downward acceleration will be less than $g$. This produces an asymmetry that is not shown in your equation.

• Great answer, addressed my question perfectly, I missed such an elementary point, of course we can't apply the suvat equations for non constant acceleration. Thanks – SNB Aug 31 '16 at 17:04

Think in terms of Energy. Object has to do some work to overcome the air friction, when it goes up and when it comes down. So the kinetic energy, potential energy interconvertion is not 100%.

When computing simple ballistic type equations, like the arch of projectile, the Newtonian equations are indeed really only a first order approximation, but that is more than adequate for studies of motion. When you want to accurately describe what a projectile flight would really be, the actual process becomes much more complex. But for study, you are looking at the idealized concept only, so you ignore things like air resistance. If you wanted to actually do something like estimate how far a struck baseball will travel given its initial angle and speed, these equations will give you the correct ideal answer, for a no air resistance and a perfect arch. In real life, that is not the actual path as there is air resistance. Next, air resistance is not a constant, or even linear compared to speed, it will be variable. The spin of the ball will cause the resistance to change. Humidity or the air while alter flight characteristics. Temperature, humidity, altitude, how the ball is deformed by being struck by the bat, uneven distribution of mass of the ball, wind gusts, if it hits a bug in flight, all these will have an affect. Some will be trivial, maybe make a 1 inch difference in landing point for a 500 foot flight, others will be very noticeable. But to correctly compute all of these variables is a much more complex process and is not needed for the basic concepts.

Now in the same scenario, if we consider air resistance, my teacher tells me that the final velocity does not equal that of the first. (Magnitude wise) But doesn't the equation of motion stated above contradict this?

That's not the equation of motion when air drag is accounted for. In that case the equation of motion is (with $y$ the vertical axis):

$$m\ddot{y}=-mg-k\dot{y}$$

(for relatively low velocities $v_y=\dot{y}$, then the drag force $F_d=-k\dot{y}$)

$$\ddot{y}+\frac{k}{m}\dot{y}+g=0$$

You can find the full solution to this differential equation below the page break. The result is:

$$\implies y=\Big(\frac{k}{m}v_0+g\Big)\Big(1-e^{-\frac{k}{m}t}\Big)-\frac{m}{k}gt$$ $$v_y(t)=\dot{y}=\frac{m}{k}\Big[\Big(\frac{k}{m}v_0+g\Big)e^{-\frac{k}{m}t}-g\Big]$$

In the absence of drag, the total flight time (up plus back down) is given by:

$$t_T=\frac{2v_0}{2}$$

And the velocity is then $v(t_T)=-v_0$.

In the case of air drag:

$$0=\Big(\frac{k}{m}v_0+g\Big)\Big(1-e^{-\frac{k}{m}t}\Big)-\frac{m}{k}gt$$

Which has two solutions, the first for $t=0\implies y=0$, the second for the time the projectile hits Earth again ($y=0$). Unfortunately that is a transcendental equation which can only be solved numerically.

The numerical solution shows that in the case of drag:

1. The total flight time is greater than $\frac{2v_0}{2}$,
2. The velocity on hitting the Earth is lower than $v_0$.

This is of course due to friction work performed by the drag force $F_d=-k\dot{y}$.

Set: $$\frac{k}{m}\dot{y}+g=u\implies \frac{k}{m}\ddot{y}=\dot{u}\implies \ddot{y}=\frac{m}{k}\dot{u}$$

So:

$$\frac{m}{k}\dot{u}+u=0$$ $$\dot{u}=-\frac{k}{m}u$$ $$\ln u= -\frac{k}{m}t+c$$ $$u=c_1e^{-\frac{k}{m}t}$$

So:

$$\frac{k}{m}\dot{y}+g=c_1e^{-\frac{k}{m}t}$$ $$\dot{y}=\frac{m}{k}[c_1e^{-\frac{k}{m}t}-g]$$ $$y=-c_1e^{-\frac{k}{m}t}-\frac{m}{k}gt+c_2$$ Initial condition 1: $$t=0, y=0$$ $$\implies 0=-c_1+c_2$$ Initial condition 2: $$t=0, \dot{y}=v_0$$ $$v_0=\frac{m}{k}(c_1-g)$$ $$c_1=\frac{k}{m}v_0+g=c_2$$ $$\implies y=\Big(\frac{k}{m}v_0+g\Big)\Big(1-e^{-\frac{k}{m}t}\Big)-\frac{m}{k}gt$$ $$\dot{y}=\frac{m}{k}\Big[\Big(\frac{k}{m}v_0+g\Big)e^{-\frac{k}{m}t}-g\Big]$$