- #1
Matt Q
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So this problem was a 2 part question. The first part goes as such.
1. A gun is fired straight up. Assuming that the air drag on the bullet varies quadratically with speed, show that the speed varies with height according to the equations:
v2 = Ae-2kx - (g/k) (upward motion)
v2 = (g/k) - Be2kx (downward motion)
in which A and B are constants of integration, g is the acceleration of gravity, and k = (c2 /m) where c2 is the drag constant and m is the mass of the bullet.
Now this problem I did not have any particular problem solving after some guidance from my professor. For this one I just took the integral of F(v) = -mg - cv2 and F(v) = mg - cv2 for the upward and downward motions respectively. It is the following question that I am hungup on.
2. Use the above result to show that, when the bullet hits the ground on its return, the speed is equal to the expression:
((v0vt) / ((v02 + vt2)^(1/2))
in which v0 is the initial upward speed and vt = (mg/c2)^(1/2) = terminal speed = (g/k)^(1/2).
I am primarily having the problem of setting up this problem. Since it wants us to take into consideration the upward and downward motion of the bullet, I would assume you would want to add the velocities of the first problem, but I could be entirely wrong about that. Also considered starting out with taking the sum of the forces of the first problem:
F(v) = -mg -cv2 and F(v) = mg-cv2 but I don't think that would work out if I integrated it then. Any help on setting up this problem would be appreciated :)
1. A gun is fired straight up. Assuming that the air drag on the bullet varies quadratically with speed, show that the speed varies with height according to the equations:
v2 = Ae-2kx - (g/k) (upward motion)
v2 = (g/k) - Be2kx (downward motion)
in which A and B are constants of integration, g is the acceleration of gravity, and k = (c2 /m) where c2 is the drag constant and m is the mass of the bullet.
Now this problem I did not have any particular problem solving after some guidance from my professor. For this one I just took the integral of F(v) = -mg - cv2 and F(v) = mg - cv2 for the upward and downward motions respectively. It is the following question that I am hungup on.
2. Use the above result to show that, when the bullet hits the ground on its return, the speed is equal to the expression:
((v0vt) / ((v02 + vt2)^(1/2))
in which v0 is the initial upward speed and vt = (mg/c2)^(1/2) = terminal speed = (g/k)^(1/2).
I am primarily having the problem of setting up this problem. Since it wants us to take into consideration the upward and downward motion of the bullet, I would assume you would want to add the velocities of the first problem, but I could be entirely wrong about that. Also considered starting out with taking the sum of the forces of the first problem:
F(v) = -mg -cv2 and F(v) = mg-cv2 but I don't think that would work out if I integrated it then. Any help on setting up this problem would be appreciated :)
