using d^2(h)/dt^2=−g+k(dh/dt)^2, find an expression for the terminal velocity in terms of k and g? g is the acceleration due to gravity, k is a constant, h(t) is the height of the falling object, dh/dt is its velocity, and d^2(h)/dt^2 is its acceleration.

Question

using d^2(h)/dt^2=−g+k(dh/dt)^2, find an expression for the terminal velocity in terms of k and g? g is the acceleration due to gravity, k is a constant, h(t) is the height of the falling object, dh/dt is its velocity, and d^2(h)/dt^2 is its acceleration.

Steve · Answer

Hmmm. velocity is dh/dt. Making that substitution, we have

v' = -g + kv
dv/(kv-g) = dt
1/k log(kv-g) = t+c
log (kv-g) = k(t + c)
kv-g = e^(k(t+c))
v = 1/k (e^(k(t+c) + g)

So, as t->infinity,
v->g/k if k<0
v->infinity if k>0

Brad · Answer

Could you explain as to how you got that answer?

Steve · Answer

where do you get lost? It's pretty basic integration and solution of exponentials.

Cody · Answer

I am stuck, too. We are trying to find an expression for terminal velocity, or where (d^2h/dt^2) = 0, in terms of g and h. I do not think we are allowed to integrate just yet, so how would we do this?

Cody · Answer

I feel really stupid, because I think the answer may just be the sqrt(g/k).

using d^2(h)/dt^2=−g+k(�dh/dt)^2, find an expression for the terminal velocity in terms of k and g? g is the acceleration due to gravity, k is a constant, h(t) is the height of the falling object, dh/dt is its velocity, and d^2(h)/dt^2 is its acceleration.