A ball bearing is dropped from rest at a point A. At theinstant it passes a mark

Question

A ball bearing is dropped from rest at a point A. At theinstant it passes a mark 4.8 m below A, another ball bearing isreleased from rest from a position 5.8 m below A. At what timeafter its release will the second ball bearing be overtaken by thefirst? How far does the second bearing fall in that time? A ball bearing is dropped from rest at a point A. At theinstant it passes a mark 4.8 m below A, another ball bearing isreleased from rest from a position 5.8 m below A. At what timeafter its release will the second ball bearing be overtaken by thefirst? How far does the second bearing fall in that time?

Explanation / Answer

xf = xo + vot +.5(9.8m/s2)t2 Use above equation to find a value for time for the firstball. xf = 4.8 xo = 0 vo = 0 Solve for t = .9897 plug this value into the equation: vf =vo + (9.8m/s2)t vo = 0 Solve for vf = 9.6994 What this now allows us to do is to "pretend" that the ball isdropped 4.8 meters below a at an intital velocity of9.6694m/s, instead of from A with a 0 initial velocity. Now we can set the two equations of motion for both ballsand set them equal to each other and solve for time (t). xo + vot+ .5(9.8m/s2)t2 = xo +vot + .5(9.8m/s2)t2
^^ firstball                        ^^second ball x0 =-1                              x0= 0     vo=9.6694m/s                 vo= 0
plugging in and solving for t: t = .10s

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