A ball is thrown upward from the ground with an initial speed of v . At the same instant, another ball is dropped from a building of height 20 ~m . If the balls are at the same height after 0.8 ~s , then the magnitude of v is (Take, acceleration due to gravity =10 ~ms ⁻² )

Step-by-step solution

The given situation is shown below Both balls meet at point C after 0.8 ~s i.e Let t₁=t₂=0.85 B C=20-h For the first ball, aligned h & =v t₁- 1 2 g t₁^2 h & =v 0.8- 1 2 10 0.8^2 aligned For the second ball, aligned 20-h & =0+ 1 2 g t₂^2 20-h & = 1 2 10 (0.8)^2 20-h & =3.2 h & =20-3.2 h & =16.8 ~m aligned Putting the value of h in Eq. (i), we get array rlrl & 16.8 & =v 0.8- 1 2 10 0.8^2 16.8 & =0.8 v-3.2 20 & =0.8 v & v & = 20 0.8 =25 ~ms ⁻¹ array