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:
- The total flight time is greater than $\frac{2v_0}{2}$,
- 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:
$$t=0, \dot{y}=v_0$$
$$v_0=\frac{m}{k}(c_1-g)$$
$$c_1=\frac{k}{m}v_0+g=c_2$$
$$\implies y=\Big(\frac{k}{m}v_0+g\Big)\Big(1-e^{-\frac{k}{m}t}\Big)-\frac{m}{k}gt$$
$$\dot{y}=\frac{m}{k}\Big[\Big(\frac{k}{m}v_0+g\Big)e^{-\frac{k}{m}t}-g\Big]$$
