CBSE Class 11-science Answered

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