a container with mass m kg is dropped by a helicopter from height h km at time t

Question

a container with mass m kg is dropped by a helicopter from height h km at time t=0, with zero velocity. from the outset, its fall is controlled by gravity and the force of air resitance, f(v)= -kv, where v is the current velocity of the container. in t seconds after the drop, a parachute opens, resulting in an increase of air resistance up to f(v) = -kv. determine the time t at which the container touches the ground. and its velocity at this moment. if m = 200 kg, h = 2000 m, t = 20 s, k = 10 kg/s, and k = 400 kg/s

Explanation / Answer

I will take the positive direction of x as upward, with x = 0 being the surface of the ground.
I will also assume that we can treat the gravitational force as being constant over the relevant distances.

The forces on the object when it is first released are gravity and the frictional drag, and the total force is the sum of these:

F = -M*g - k*v(t)

From Newton's second law, F = M*a = M*x'' = M*v', and v(t) = x', so:
M*v' = -M*g - k*v
v' = -(k/M)*(v + g*M/k)

This is a separable equation:
dv/(v + g*M/k) = -(k/M) dt
ln(v + g*M/k) = -k*t/M + c
where c is the constant of integration.

v = C*exp(-k*t/M) - g*M/k
where C = exp(c) is just another way of writing the constant.

We know that initially, v(0) = 0, so:
0 = C - g*M/k
C = g*M/k

v(t) = (g*M/k)*(exp(-k*t/M) - 1)

Now, v = x', so:
x' = (g*M/k)*(exp(-k*t/M) - 1)
dx = (g*M/k)*(exp(-k*t/M) - 1) dt
x = -(g*M/k)*[(M/k)*exp(-k*t/M) + t] + d
where d is the second constant of integration.

Taking x(0) = H, we have:
H = -(g*M^2)/k^2 + d
d = H + (g*M^2)/k^2
x(t) = H + ((g*M^2)/k^2)*(1 - exp(-k*t/M)) - g*M*t/k

Plugging in the appropriate values for H, g, M, and k (noting that k must have units of mass/time, so I assume k = 10 kg/sec), we get:
v(t) = (9.8*200/10)*(exp(-10*t/200) - 1)
v(t) = (196 m/s)*(exp(-t/20sec) - 1)

x(t) = 2000 m + ((9.8*200^2)/10^2)*(1 - exp(-10*t/200)) - 9.8*200*t/10
x(t) = 2000 m + (3920 m)*(1 - exp(-t/20sec)) - t*196 m/s)

At t = 20 sec,
v(20 sec) = 26.52.0 m/s
x(t) = 2000 m + 2478 - 3920
x(20 sec) = 558 m

Now we can simply "restart the clock" when the parachute opens, This time, with an initial height of 558 m, and an initial velocity of 26.52 m/s. In addition, the frictional force is increased to a new value of 400 kg/sec because of the parachute.

Going back to the solution for the velocity, we have:
v = C*exp(-K*t/M) - g*M/K
Plugging in the values once the parachute opens:
v(t) = C*exp(-t*400/200) - 9.8*200/400
v(t) = C*exp(-2t) - 4.9
v(0) = -4.9 m/s
Thus C = -4.9 m/s
v(t) = -4.9*exp(-2t) - 4.9

Now integrate this to get the position as a function of time:
x(t) = -(M/K)*C*exp(-k*t/M) - g*M*t/K + d
where d is yet another constant of integration.
Again plugging in the values of the constants:
x(t) = -(200/400)*(-4.9)*exp(-400*t/200) - 9.8*200*t/400 + d
x(t) = 2.45*exp(-2t) - 4.9t + d
Now x(0) = 558 m, so:
558 = 2.45 + d
d = 555.55 m

x(t) = 2.45*exp(-2t) - 4.9t + 555.55

Solve it to find t and then v(t)

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