What is the height when the ball is thrown vertically upward with air resistance??

Question

A ball is thrown vertically upward with u velocity. There is air resistance and the air resistance is directly proportional to square of ball's velocity,u. Find the height which the ball can reach. I started with, $$a=-g-{{k\over m}u^2},~~~~~~~~~ {d^2x\over dt^2} = -g - {k\over m}\left({dx\over dt}\right)^2$$

Is it true? and How can I solve this differential equation?

Answer 1

HINT...since you are looking for the height not the time, write $$a=v\frac{dv}{dx}=-(g+\frac kmv^2)$$

Answer 2

HINT

Yes your set up is correct indeed we have that

• $a=-g-\frac k m \cdot v^2 \implies \frac{dv}{dt}=-g-\frac k m v^2 \implies \frac{v'(t)}{-g-\frac k m v^2}dt=dt$

which can be integrated (see here for the details).