Solving Projectile Motion: Finding Max Height & Final Speed

[Gogsey]

For a force F = −mg ± cv^2 (the sign is chosen so the drag force is opposite the motion), you can use a trick: let u = v^2, and generate a differential equation for u(x) (not u(t)!), which is not too hard to solve. (Hint: the chain rule gives this dv/dt=dv/dx * dx/dt should help.)
a) Solve this to show that a projectile thrown vertically upwards from the ground with speed v0 reaches a maximum height y=V(t)^2/2g*(ln(1 + V(o)^2/V(t)^2, where V(t) is the terminal speed.
b) Show that the projectile hits the ground with speed V(f) given by:
V(f)^-2=V(o)^-2 + V(t)^-2

So, part a) is ok, I have that after quite a bit of math, but I'm a little lost on b). When a projectile hits the ground, that's usually at t((y) = 0, but there is not time term in this equation.

 

[CompuChip]

So let's call T the time when the projectile hits the ground.
What do you know about the value of y(T)?
The final velocity is V(T) = V(f).
This should give you an equation.

Also, please check the answer you are supposed to derive... is there really a V(t) in there?

 

[gabbagabbahey]

Also, please check the answer you are supposed to derive... is there really a V(t) in there?

I obtained the desired result, so I suspect that you may have used the wrong sign in front of 'c' during the falling portion of the motion.

 

[gabbagabbahey]

So, part a) is ok, I have that after quite a bit of math, but I'm a little lost on b). When a projectile hits the ground, that's usually at t((y) = 0, but there is not time term in this equation.

I hope you aren't going to all the trouble of actually finding x(t) (or y(t) if that's what you're calling it)!

All you really need to do is find $u(x)=v^2(x)$ as the projectile falls from a height of $$x=\frac{v_t^2}{2g}\ln\left(1+\frac{v_0^2}{v_t^2}\right)$$ to $x=0$. The impact speed will simply be $v_f=v(x=0)=\sqrt{u(x=0)}$