CBSE Class 11-science Answered

Ball dropped from the tower :

Initial position of the ball   x_o1 = 200m

Initial velocity of the ball   u₁   = 0

Acceleration                       a₁ = g = -10 m/s²

Therefore, the position of the ball at any instant

x₁ =   x_o1  + u₁t + a₁ t² /2 =    200   - 5 t²...............................(i)

Ball projected from ground :

Initial position of the ball   x_o2 = 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 =   x_o2 + 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.