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.

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

Could you explain as to how you got that answer?

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

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?

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

To find the expression for the terminal velocity, we need to find the point at which the acceleration becomes zero. At terminal velocity, the object reaches a constant velocity and its acceleration is zero.

Given the equation: d^2(h)/dt^2 = -g + k(dh/dt)^2

Setting the acceleration equal to zero, we have:

0 = -g + k(dh/dt)^2

Rearranging the equation, we get:

g = k(dh/dt)^2

Now, we can solve for dh/dt (velocity):

(dh/dt)^2 = g/k

Taking the square root of both sides, we have:

dh/dt = sqrt(g/k)

Therefore, the expression for the terminal velocity (v_t) in terms of k and g is:

v_t = sqrt(g/k)