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.)(adsbygoogle = window.adsbygoogle || []).push({});

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, thats usually at t((y) = 0, but there is not time term in this equation.

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# Homework Help: Quadratic Drag

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