Using diff eq:

a bullet weighing 1 oz is fired vertically downward from a stationary hellicopter with a muzzle velocity of 1200 ft/sec. The air resistance (in pounds) is numericallly equal to 16^-5 v^2, where v is the velocity of the bullet in ft/sec.

a) find the velocity of the bullet as a function of time
b) if the helicopter is 3,000 ft high, how long does it take for the bullet t o hit the ground?

Question

Using diff eq:

a bullet weighing 1 oz is fired vertically downward from a stationary hellicopter with a muzzle velocity of 1200 ft/sec. The air resistance (in pounds) is numericallly equal to 16^-5 v^2, where v is the velocity of the bullet in ft/sec.

a) find the velocity of the bullet as a function of time
b) if the helicopter is 3,000 ft high, how long does it take for the bullet t o hit the ground?

hba · Answer

i fear this is a physics question

anonymous · Answer

is for my differential equation class ... I started by doing m* dv/dt = F1+F2   but the integral I got with respect to v was to complicated to solve and that's where im stuck

hba · Answer

ok wait

turingtest · Answer

So the air resistance is
F=16^(-5v^2)
???????

jamesj · Answer

Yes, so ma = m v' = F = -mg + K v^2, where K is the constant.


Hence dv/dt = -g + K/m v^2.  Write a^2 = K/(mg), then
dv/(a^2 v^2 - 1) = g dt

Now integrate (using partial fraction, etc.)

jamesj · Answer

Then apply the initial conditions