A kangaroo can jump 2.5 m vertically. What must be the initial velocity of the kangaroo?

Explanation

`a = 9.8 m//s^(2)`
`s= 2.5 m`
`v= 0`
u= ?
`v^(2)=u^(2)+2as`
`(0)^(2) =u^(2)+2xx(-9.8)(2.5)` Negative sign is used as the acceleration is in the direc- tion opposite to that of velocity.
`0 =u^(2)-49`
`u^(2)=49`
`u=7 m//s`