A body dropped from top of a tower falls through ` 40 m` during the last two seconds of its fall. The height of tower in m is ( g= 10 m//s^@)`

Options

A. 60 m
B. 45 m
C. 80 m
D. 50 m

Answer

B

Explanation

Let the body fall through the height of tower in t seconds.
From , `D_(n) = u +g/2(2n-1)` we have ,
total distance travelled in last 2 second of fall is
`D=D_(t)+D_(t-1)=[0+g/2(2t-1)]+[0+g/2(2(t-1)-1)] `
`=g/2 (2t -1) +g/2 (2t-3) =g/2(4t-4)=10/2xx4(t-1)`
or `40 = 20 (t-1) " or "t = 2 +1 =3s`
Distance travelled in t seconds is
`s = ut =1/2 "gt"^2 =0 +1/2 xx 10 xx 3^(2) =45m`
The height of the tower is 45 m.