Question

A ball is dropped from the top of a tower. After 2 s another ball is thrown vertically downwards with a speed of `40 ms^(-1)`. After how much time and at what distance below the top of tower the balls meet ?

Answer 1

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.

For first ball, `h=(1)/(2)g t^(2)` …..(i)
For second ball, `h=40 (t-2)+(1)/(2)g(t-2)^(2)` .....(ii)
Eqs. (i) and (ii), we get
`40(t-2)+(1)/(2)g(t-2)^(2)=(1)/(2)g t^(2)`
`40(t-2)=(1)/(2)g[t^(2)-(t-2)^(2)]`
`40(t-2)=(1)/(2)xx10(2t-2)xx2`
`4t-8=2t-2 rArr t=3 s`
Distance below the top of tower the balls meet
`h=(1)/(2)g t^(2)=(1)/(2)xx10xx3^(2)=45 m`
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.
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.