A ball is dropped from the top of a building after 2 seconds

Text Solution

Solution : Let the balls meet at distance h below the top of water t second after dropping of first ball. The second ball takes time (t - 2) seconds. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NEET_PHY_OBJ_V01_C03_S01_045_S01.png" width="80%"> <br> For first ball, `h=(1)/(2)g t^(2)` …..(i) <br> For second ball, `h=40 (t-2)+(1)/(2)g(t-2)^(2)` .....(ii) <br> Eqs. (i) and (ii), we get <br> `40(t-2)+(1)/(2)g(t-2)^(2)=(1)/(2)g t^(2)` <br> `40(t-2)=(1)/(2)g[t^(2)-(t-2)^(2)]` <br> `40(t-2)=(1)/(2)xx10(2t-2)xx2` <br> `4t-8=2t-2 rArr t=3 s` <br> Distance below the top of tower the balls meet <br> `h=(1)/(2)g t^(2)=(1)/(2)xx10xx3^(2)=45 m` <br> Note : 1. By boundary condition, we mean that velocity or displacement at some time (usually at t = 0) should be known to us. Otherwise we cannot find constant of integration. <br> 2. Equation a = v dv / ds or v dv = a ds is useful when acceleration displacement equation is known and velocity displacemnt equation is required.