A ball thrown up from the ground reaches a maximum height of 20m Find :
Its initial velocity,
the time taken to reach the highest point,
its velocity just before hitting the ground ,
its displacement between `0.5s"and"2.5s`,
time when displacement is 15m

Explanation

Using `v^(2)=u^(2)+2as ` for upward motion , we get
`0^(2)=u^(2)+2(-g)(+20) rArr u=19.8m//s`.
`t=((v-u))/a=(0-19.8)/(-9.8)=2.02s`
For the complete up - down trip.
`v^(2)=u^(2)+2a(0)rArr v^(2)u^(2)rArrv=-u=-19.8m//s`
Height at `t=0.5s "is" y_(1)=19.8(0.5)-4.9(0.5)^(2)=8.68m`.
Height at `t=2.5s "is " y_(2)=19.8(2.5)-4.9(2.5)^(2)=18.9m`
`:.` Displacement `=y_(2)-y_(1)`
`=18.9m-8.68m=+10.2m`.
`15=19.8t-4.91t^(2)rArrt=1.01s, 3.03s.`
at `t=1.01s.` ball is going up and at `t=3.03s,` it is coming down .