A ball is thrown vertically upward with velocity `20 m//s` from a tower of height 60 m. After how much time and with what velocity will it strike the ground?

Explanation

Method I: We will breah the motion in two parts.
`O` to `A` (vertically upward motion):
At the highest point, velocity=0
`v=u-g t`
`0=20-10t_(1)impliest_(1)=2 s`
`v^(2)=u^(2)-2gh`
`0=(20)^(2)-2xx10xxhimpliesh=20 m`
`A` to `B` (vertically downward direction):
`AB=h+60=20+60=80 m`
`80=0+1/2g t_(2)^(2)=5t_(2)^(2)impliest_(2)=4 s`
The ball will strike the ground after `6 s`.
`v_(0)=u+g t_(2)=0+10xx4=40 m//s`
Note: The above method is lengthy. We will not break motion in part unless necessary. We know that equations of motion are applicable for displacement.
Method II: Now taking `O` as the origin and vertically upward direction positive, from `O` to `B` through A, the displacement is negative, i.e.
`h=-60m`
`h=ut-1/2 g t^(2)(u uarr g darr)`
`-60-=20t-5t^(2)`
`t^(2)-4t-12=0`
`(t-6)(t+2)=0`
`t=6 s, t=-2 s`
Time `t=-2 s` is not possible, time cannot be-ve.
The ball will steike the ground after `6 s`.
`v=u-g t`
`v_(0)=20-10xx6=-40 m//s`
Negative sign indicates that the velocity is in downward direction.