(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.

(i) Write down the equation of a freely falling body under gravity. (

1.	(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.
Answer» Solution :`(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`