The problem statement is as follows:
Two balls of masses $M$ and $2M$ are thrown horizontally with the same initial velocity $u$ from the top of a tall tower and experience a viscous drag of $-kv$ ($k>0$) where $v$ is the instantaneous velocity. Compare the ranges of the two projectiles.
Now, I separately considered the motion of a ball horizontally and vertically. The only component of acceleration along the horizontal direction was provided by drag. So the equation would be , a = -kp/m( a is the horizontal acceleration, and p is the instantaneous horizontal velocity. ) The acceleration is inversely proportional to the mass and hence the heavier ball would have lesser acceleration, and then the heavier ball would hit the ground further away. My instincts directed me to integrate twice as follows: p dp/dx = -kp/m ( x is the horizontally displacement) and then integrating w.r.t time(t), this is the equation I obtained: ln(x) = -kt/m. Now I am second guessing myself because I have a feeling that the time of flight wouldn't be equal for both the masses. I tried to calculate the time of flight by analysing the vertical motion and obtained a differential equation I am finding hard to solve.(I'm just a high school student who's not yet done with half of the course's calculus part :P) Was my first solution correct? Or will successive integration give me a different answer altogether?