`(i)` Write down the equation of a freely falling body under gravity.
`(ii)` A ball is thrown vertically upwards with the speed of `19.6ms^(-1)` from the top of a building and reaches the earth in `6s`. Find the height of the building.

Explanation

`(i)` Equation of motion are
`(i) v=u+at`
`(ii) s=ut+(1)/(2)at2`
`(iii) v^(2)=u^(2)+2as`
If `a=g`, `s=y`
then `v=u+''gt''`
`y=ut+(1)/(2)"gt"^(2)`
`v^(2)=u^(2)+2gy`
if `u=0`, then `v="gt"`
`y=(1)/(2)"gt"^(2)`
`v^(2)=2gy`
If `t=T`, `y=h` then
`h=(1)/(2)"gt"^(2)`
`T^(2)=(2h)/(g)impliesT=sqrt((2h)/(g))`
`v^(2)=2gh`
[if `v=v_(g)`] `v_(g)=sqrt(2gh)` where `v_(g)` speed of the body after reaching ground.
`(ii)` Let `h` height of the building let the ball attain height `h'` above the builiding.
At `h'` the velocity `v=0`

By applying equation of motion `v^(2)-u^(2)=2gh`
`0^(2)=(19.6)^(2)-2gh`
`2gh'=(19.6)^(2)`
`h'=(19.6xx19.6)/(2xx9.8)`
`h'=19.6m`
Times taken by the ball to reach `h'` is `t'` (say)
`v=u+at|a-g,t-t'|`
`0=19.6-"gt" '`
`t'=(19.6)/(9.8)-2s`
Time taken by the ball to fall from height `(h+h')`
`=6S-2S=4S`
We know that, `S=ut+(1)/(2)"gt"^(2)`
i.e. `(h+h')=ut+(1)/(2)"gt"^(2)`
here `u=0`
So, `h+h'=(1)/(2)"gt"^(2)`
`h+19.6=(1)/(2)xx9.8xx(4)^(2)`
`h=9.8xx8-19.6`
`h=78.4-19.6`
height of the building `h=58.8m`