# Thread: Newtonian Mechanics 1

1. ## Newtonian Mechanics 1

A parachutist whose weight is 75 kg drops from a helicopter hovering 2000 m above the ground and falls towards the ground under the influence of gravity. Assume that the force due to the air resistance is proportional to the velocity of the parachutist, with the proportionality constant k1 = 30 kg/sec when the chute is close and k2 = 90 kg/sec when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will it be before the parachutist reach the ground? ( Take g = 9.81 m/sec2.)

2. Originally Posted by bobey
A parachutist whose weight is 75 kg drops from a helicopter hovering 2000 m above the ground and falls towards the ground under the influence of gravity. Assume that the force due to the air resistance is proportional to the velocity of the parachutist, with the proportionality constant k1 = 30 kg/sec when the chute is close and k2 = 90 kg/sec when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will it be before the parachutist reach the ground? ( Take g = 9.81 m/sec2.)
First find the time and height when the velocity reaches -20 m/s (here we take height x positive upwards) with the parachute closed. Then use the height time and speed as initial conditions for the final decent with the parachute open.

CB

3. Originally Posted by bobey
A parachutist whose weight is 75 kg drops from a helicopter hovering 2000 m above the ground and falls towards the ground under the influence of gravity. Assume that the force due to the air resistance is proportional to the velocity of the parachutist, with the proportionality constant k1 = 30 kg/sec when the chute is close and k2 = 90 kg/sec when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will it be before the parachutist reach the ground? ( Take g = 9.81 m/sec2.)

The equation of motion of the parachutist falling is :

$m\frac{dv}{dt} = mg-k_{1}v$.................................................. ....................................1

I hope you can derive that by applying the forces acting on the parachutist

From (1) we have : $\frac{mdv}{mg-k_{1}v} =dt$

And integrating from t=0 to t= $t_{1}$ and from v=0 to v= 20m/sec we have :

$t_{1} =\frac{m}{k_{1}}ln\frac{mg}{mg-20k_{1}}$

And substituting the values given by the problem we find the time taken before the parachut opens .

And since the equation for the motion is the same for the part where the parachut is open ,in a similar way you can calculate the time taken for this part ,and hence the overall time taken before the parachutist reach the ground