Posted by kgaogelo on Wednesday, March 13, 2013 at 2:46pm.
a bot throws a cricket ball witg an initial velocity of 5m.s from the edge of a tall building.it passes the edge of the building on its way down and strikes the ground below after 5seconds (neglect effects of friction. question1...how does the acceleration of the ball moving downwards compare with the upwards movement?calculate the time taken react the higher point.calculate the height the building.draw a position vs time graph inserting all known values.

physical science  Henry, Friday, March 15, 2013 at 1:00pm
V^2 = Vo^2 + 2g*h
h = (V^2Vo^2)/2g
h = (025)/19.6 = 1.28 m. = Max. ht.
above bldg.
V = Vo + g*t
Tr = (VVo)/g = (05)/9.8 = 0.51 s. =
Rise time.
Tf1 = Tr = 0.51 s. = Time to fall back to edge of bldg.
Tf2 = 5 =  Tr  Tf1=50.51  0.51=3.98 s.
Hb = Vo*t + 0.5a*t^2
Hb = 5*3.98 + 4.9*(3.98)^2 = 97.5 m. =
ht. of bldg.

physical science  Henry, Friday, March 15, 2013 at 3:50pm
T = Time in seconds.
h = Ht above gnd. in meters.
(T,h).
(0,97.5)
(0.51,98.8)
(1.01,97.5)
(5,0).
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