PeroK said:
I would take this approach to this problem. Physics and mathematics is all about being precise, and systematic.
I'm all for that, so let's do some math and, while we're a it, put in some (realistic) numbers. I will model the air resistance as proportional to the velocity. With "down" taken to be the positive direction, Newton's second law gives $$\frac{dv}{dt}=-\frac{b}{m}v+g ~~~~(b>0).$$The general solution of this is $$v(t)=v(0)e^{-(b/m)t}+\frac{mg}{b}\left[1-e^{-(b/m)t}\right].\tag{1}$$Let ##t=t_o## be the deployment time when the parachute opens. We assume that deployment is instantaneous. We also assume that we have two air resistance coefficients, ##b_1## and ##b_2## before and after deployment respectively. Finally, assuming that the parachutist starts with zero vertical velocity and falls straight down, we write
$$v_1(t)=\frac{mg}{b_1}\left[1-e^{-(b_1/m)t}\right]~~~~~~(t\leq t_o).$$ The net force ##F## at deployment mentioned in the problem is $$F=m \left . \frac{dv}{dt}\right |_{t=to}=mg~e^{-(b_1/m)t_o}\implies e^{-(b_1/m)t_o}=\frac{F}{mg} .$$ At the time of deployment, $$v(t_o)=\frac{mg}{b_1}\left[1-e^{-(b_1/m)t_o}\right]=\frac{mg}{b_1}\left[1-\frac{F}{mg}\right].$$ Now the velocity is continuous through the deployment and we use the expression above as the initial velocity in equation (1) to write the velocity after deployment: $$v_2(t) =\frac{mg}{b_1}\left[1-\frac{F}{mg}\right]e^{-(b_2/m)t}+\frac{mg}{b_2}\left[1-e^{-(b_2/m)t}\right]~~~~~~(t>t_o)$$ Of interest is the acceleration after deployment which, after some algebra, is given by$$a_2(t)=\frac{dv_2}{dt}=g\left[1+\frac{F}{mg}-\frac{b_2}{b_1}\right]e^{-(b_2/m)t}.$$According to our adopted convention, if the factor multiplying the exponential is positive the acceleration is down, else it is up. We are told by the problem that ##F## is "somewhat less" than the weight. This means that ##1< (1+\frac{F}{mg})< 2.## We also know that ##b_2 > b_1## because the air resistance increases when the parachute opens. The terminal velocity is given by ##v_{\text{ter}}=\dfrac{mg}{b}##, therefore $$\frac{b_2}{b_1}=\frac{v_{\text{ter,1}}}{v_{\text{ter,2}}}.$$ The web tells me
here that in the stable belly-to-earth position the terminal velocity (parachute not deployed) is ##v_{\text{ter,1}}=~##120 mph. It also tells me
here that the landing velocity of an open parachute is ##v_{\text{ter,2}}=~##15 mph. This means that the ratio ##b_2/b_1=v_{\text{ter,1}}/v_{\text{ter,2}}=8## which makes the factor multiplying the exponential
negative. Thus, the acceleration and hence the net force are directed "up" after deployment. In order for the net force to be down, one would need a landing speed that is in excess of 60 mph. Ouch!
