Physics mechanics help?

If you assume that the horizontal velocity is constant ( no air resistance) and that there is no initial vertical component ( dropped ) then h = 1/2 g t^2 -> t = sqrt( 2h/g)

and the horizontal distance moved is vt = v*sqrt( 2h/g) = 30 * sqrt( 2* 300 /9.8)  Which is about 235 m.  But if the package is moving at 30 m/s when it hits the ground it is going to bounce or roll some unknown distance.  So "about 200m" is probably the best answer that could be given.

Since the force of gravity only acts in the vertical direction, the supplies move horizontally at the same speed as the plane.  Since the plane is not moving vertically, the packages have 0 speed in the vertical direction when first released.  So we can copute teh time it take them to fall h = 300m:

 y = -1/2 gt^2    "-" = down,  "+" = up

Set y = 300 m and g = 9.8 m/s^2  --->  t = sqrt(2y/g) = sqrt(600/9.8) m = 7.825 s

The package move x meters along the ground in th etime t = 7.825 s :

x = v*t   where v = 30 m/s --->   x = 234.74 m

So the drop has to be made while the plane is 234.7 m away from the survivors

Sorry, my answer is "Anonymous" as your post is Anonymous. There is no need for an Anonymous question in this category, Most 'Anonymous' posts are trolls, maybe yours is not, but the high likelihood is there. If you post as yourself, I would be glad to help you with any problem you have.

I spend a lot of time on each problem, I just want to make sure it's for a worthwhile cause.