If a bowling ball falls out of an airplane going 500 mph at 30,000 feet, how long til it hits ground.?

Question

Assuming ground is at sea level and no weather factors or anything. Standard conditions.

Answer 1

x= -16t^2 + xo
0 = -16t^2 + 30000
-30,000 = -16t^2
t= sqr(-30,000/-16)
t= 43.3s

500mph has no effect on falling object because the speed in horizonal.

Answer 2

It comes right down to which will go in the route of the finest suitable consequences from air resistance. the obtrusive answer is the sea coast ball because it really is more effective (more effective floor section), consequently is subject to larger frictional forces, which wll have an greater result because the sea coast ball weighs a lot less. yet that assume the frictional forces are purely dependant on the outdoors are of the ball. a large adjustments is the bowling ball has finger holes. those fingerholes may supply upward push to turbulant air bypass around the bowling ball with will sluggish the bowling ball down more effective than the sea coast ball. word I reported CAN. it really is dependant on the orientation of the finger holes, and the quantity and route of spin on the ball even as it really is dropped. If the bowling ball is dropped with the fingerhole element up, and the spin is such that they stay pointing up (no spin or spinning like a good strickly via the vertical axis) then the fingerhole received't result the air bypass and the bowling ball will hit the floor first, in the different case the sea coast ball will hit the floor first.

Answer 3

The airplane's speed is irrelevant.
s=1/2 att
30000=1/2 32ft/s/s tt
30000=16ft/s/s tt
tt=30000/16
tt=1875
t=43.3 sec
Assuming no air resistance