Using the information given in example 2, find (i) the greatest height reached by the stone above the ground.
(ii) the velocity with which it strikes the ground
(iii) time taken to reach the maximum height

Explanation

Let.s be the height covered in upward motion then
`v^(2)=u^(2)+2gs`
`0=(14.7)^(2)+2xx(-0.8)s`
`rArr s=((14.7)^(2))/(19.6)=11.025m`
`therefore` The greatest height above the ground =s+s
=11.025+196
=207.025m
(ii) For the downward motion of the stone,
u=0,g=`9.8m//s^(2)`, s=207,028m
Using `v^(2)=u^(2)+2gs`
`=0+2xx(9.8)xx207.025`
`rArr v=63.7m//s`
(iii) Let t be the time taken to reach the maximum height. Then, for the upward motion
`u=14.7m//s, v=0g=-9.8m//s^(2)`
Using v=u+gt, we get
`t=(v-u)/(g)=(0-14.7)/(-9.8)=1.5s`