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

Answer

B

Explanation

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