Quote: If dust falls at approximately .5meters per second off a 200ft high wall. the wind is blowing at a 45 degree angle up with a force of 1 meter an hour. when does the dust hit the ground?
The point of this problem is to find out the resulting downwards velocity of the dust given that the wind is blowing at an upward angle.
Note that since we are not told the mass of the dust it is assumed we can neglect the force of gravity for this problem and that the particles velocity will be constant.
The velocity upwards that the wind gives the dust is 1 m/h * sin(45 degrees) = sqrt(2)/2 m/h = sqrt(2)/7200 m/s; (3600 seconds to an hour)
The upwards velocities just add so the resulting velocity is
0.5 + sqrt(2)/7200 m/s = (3600 + sqrt(2))/7200 m/s ~ 0.5 m/s; (note the wind has almost no effect, are you sure it wasn't 1 m/s windspeed?)
The time it will take to fall 200 ft is 200 feet divided by the velocity (in feet/second, we'll have to convert from m/s to ft/s)
There are about 3.05 feet/meter so the velocity in feet/s is 0.5m/s * 3.05 ft/m = 1.525 ft/s
So the time is now given by 200ft/1.525 ft/s = 131.15 seconds