from the top of a tower 100m in height a ball is dropped and at the same time another ball is

Asked by  | 23rd Jul, 2009, 02:17: PM

Expert Answer:

Ball dropped from the top of tower :

Initial position of ball  x01 = 100 m

Initial velocity of ball  u1  = 0

Acceleration                a1  = g = -10 m/s2

so, the position of ball at any instant 

x1  =  x01  + u1 t + a1 t2 /2   =  100 - 5 t2 ...........................................(i)

Ball projected from ground :

Initial position of ball  x02 =  0

Initial velocity of ball  u2  =   25 m/s

Acceleration                a2  = g = -10 m/s2

so, the position of ball at any instant 

x2  =  x02  + u2 t + a2 t2 /2   =  25t - 5 t2 ...........................................(ii)

When the balls will meet the position of both the balls will be same

i.e. x1 = x2

or,  100 - 5 t2  = 25t - 5 t2

or, t = 4 s

substitute t in equn (i)

x1  =   100 - 5 t2 = 100 - 5 (4)2 = 20 m

Thus, the balls meet at t = 4s at a height of 20 m from the ground

Answered by  | 23rd Jul, 2009, 08:05: PM

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