A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`.

Explanation

If the body falls a height h in time t, then `" "h=1/2 g t^(2)`
`[u=0 " as the body starts from rest"]` …(i)
Now, from Equation (i) and (ii) distance travelled in the last second.
`h-h'=1/2g t^(2)-1/2g(t-1)^(2)` i.e., `h-h'=1/2 g(2t-1)`
But according to given problem as `(h-h')=h/2`
`rArr (1/2)h=(1/2)g(2t-1)` or `(1/2)g t^(2)=g(2t-1)` [as from equation (i) `h=(1/2)g t^(2)`]
`rArr t^(2)-4t+2=0" "` or `" " t=[4+-sqrt((4^(2)-4xx2)"]"//2)rArr t,=2+-sqrt(2)rArr t=0.59s` or `" "3.41 s`
0.59 s is physically unacceptable as it gives the total time t taken by the body to reach ground lesser than one sec while according to the given problem time of motion must be greater than `1s`.
so `t=3.41 s` and `h=1/2xx(9.8)xx(3.41)^(2)=57 m`