[Experts only: 4th 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?

Answer 1

Initial velocity, angle, range in the north-south direction are irrelevant.

That's the beauty of orthogonal coordinate systems!

Range E-W = 40 m
Velocity E-W = 12 m/s
Time in the air = 40/12 = 3 1/3 seconds

Edit ...
I should point out the problem with the above posts. A cannon ball fired with a speed of 200 m/s at an angle of 15 degrees would have a range on level ground of 2040 m. Since it's only 1200 m, we must be firing uphill. This complicates things if you try to work in the N-S up/down coordinate plane.

Answer 2

The wind did not affect the speed in the north-south direction. This speed was the horizontal vector at the moment of firing

Vns = cos(15) x 200 m/s = 193.2 m/s

The cannon ball crossed exactly 1200 meters in the north-south dorection since it fell due east of the target. Therefore it spent

1200m / 193.2m/s = 6.2s
_______
Note: the previous answer assumes that both the cannon and target are at the same elevation which is at all obvious. As a matter of fact, since he got a different answer than me proves that the elevation are different.

Answer 3

This is how I usually approach these.

For this kind of scenario, you really only need to use the time/dist/rate eq'n: d = rt

r = 200 m/s

The drift was 40 m (east). If you draw a triangle and use the pyth. thrm you'll see that the distance traveled is d = 1200.67 m.

d = rt

==> t = d/r = 1200.67/200 ≈ 6 s


t ≈ 6 s

The time to traverse the horizontal distance is not dependent on gravity.....there's no acceleration along the 'x' direction, so you can afford yourself this simplification. The time along 'x' HAS to be the same as it is along 'y', otherwise there's big problems. Unless there's more to this than what I'm seeing, I don't see how the other info is pertinent to this calculation.

Answer 4

grrr..sorry disregard i was using the wrong trajectory equations...heres the right ones...the answer is 6.21 seconds...based on this equation:

t = d/v((cos)a) where d is distance and a is the angle...my fault :)

note to below: nope i was using the wrong traj equations :) got the right ones now :)