please help.....

Asked by  | 26th Jul, 2009, 01:38: PM

Expert Answer:

Ball dropped from the tower :

Initial position of the ball   xo1 = 200m

Initial velocity of the ball   u1   = 0

Acceleration                       a1 = g = -10 m/s2

Therefore, the position of the ball at any instant

x1 =   xo1  + u1t + a1 t2 /2 =    200   - 5 t2...............................(i)

Ball projected from ground :

Initial position of the ball   xo2 = 0

Initial velocity of the ball   u2   =  50 m/s

Acceleration                       a2 = g = -10 m/s2

Therefore, the position of the ball at any instant

x2 =   xo2 + u2t + a2 t2 /2 =    50t  - 5 t2...............................(i)

When the two balls will meet , the position of the two balls will be same

i.e. x1 = x2  

or, 200  - 5 t2 50t  - 5 t2

0r, t = 200/50 = 4 s.

Substituting t = 4 n eqn (i)

x1  = 200 - 5 ( 4)2 = 120 m

Thus the balls meet at t = 4s  at a height 120 m from the ground.

Answered by  | 26th Jul, 2009, 03:51: PM

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