A ball is dropped from the top of an 80 m high tower After 2 s another ball is thrown downwards from the tower. Both the balls reach the ground simultaneously. The initial speed of the second ball is

Explanation

First ball: `h=1/2 g t^(2)`
`80=1/2xx10xxt^(2)impliest=4 s`
Second ball: Let the second ball be thrown with speed `v_(0)`
`h=v_(0)(t-2)+1/2g(t-2)^(2)`
`80=v_(0)(4-2)+1/2xx10(4-2)^(2)`
`=2v_(0)+20`
`v_(0)=30 m//s`