[Experts only: 5th repost] How much time did cannon ball spend in the air?

Question

Cannon is aimed right at the target located d = 1200m due north
from the cannon and fired. Initial speed of the ball v = 200m/s,
angle of inclination α = 15 degrees to the horizon. The ball misses
the target by r = 40m to due east, because gunmen did not
take into account the wind blowing from due west with the
speed u = 12 m/s.

How much time did the cannon ball spend in the air?


Since it is fifth repost a hint would be appropriate:
There exists a frame of reference where
the force of gravity, the force of air resistance,
and velocity of the ball are all co-planar.

'horrendous differential mess'

The solution of this problem requires no knowledge of calculus. Simple Galilean relativity (all inertial frames are equal)
and symmetry considerations is all it
takes.

Answer 1

The total flight time of the cannonball is 9,548 sec.

Let Vx(t) the ball speed in the wind direction and Vy(t) the ball speed aimed to the target.
At the time zero Vx=0 and Vy=200*cos(15)

Vx(t)=12{1-e^(-k*t)} where k is a factor of the air resistance.
From the Vx(t) we can get easily the function of shift:
Sx(t)=12*t+(12/k)*e^(-k*t) which is equal to 40m at the end of the flight

Vy(t)=200*cos(15)*e^(-k*t)
Sy(t)=(-193,185/k)*e^(-k*t) which is equal to 1200,666m at the end of the flight. It comes from (1200^2+40^2)^0,5 Pithagoras
Let use A={e^(-k*t)}/k parameter to implement and solve the two equation.

(Do not be disturb by the nonsense of negative distance, it is only because of the vectors of velocities and shifts are perpendicular to each other.)

Answer 2

The vertical component of the cannon's muzzle velocity is

200 Sin(15) m/sec, or about 51.76 m/sec

Since the wind is presumed to be parallel to the ground, and no details are given about the shape of the projectile in which to try to work out a possible vertical component of the resultant force of the wind on the projectile, nor any figures are given for air resistance, this vertical component is presumed to be independent of the projectile's flight. We compute the time it takes to be reduced to 0 by gravity:

51.76 m/sec) / (9.8 m/sec²) = 5.28 sec

Since what goes up has to come down, we have the total air time of

2 * 5.28 = 10.56 seconds.

To say that the air time equals (40 m) / (12 m/sec) = 3.33 seconds is assuming that there is no air flow around the projectile, not a good assumption to make.

Addendum: However, it's an interesting complication to suggest that we can infer the effect of air resistence on the projectile from the fact it was deflected 40 m by a 12 m/sec wind, but I need more time to think about this. It's not so simple.

Addendum 2: Yes, I can figure in the drag from air resistance as evidenced by the 40m deflection by a 12 m/sec wind, but it's an horrendous differential mess. I'd like to hear from you first before I fully work this out. .

Answer 3

The time of flight is the E-W drift divided by the wind speed plus the range through the air divided by the muzzle velocity.