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

Answer

A

Explanation

`S_(nth) = u +(a)/(2) (2n-1) ` . Total distance travelled in the last 2 seconds of fall .
` S = S_n +S_(n-1)`
`= [0+ (g)/(2) (2n-1) ] +[0 +(g)/(2) xx 2 (n-1) -1]`
` = (g)/(2) (2n-1) +(g)/(2) (2n-3) `
` = (g)/(2) (2n-1) +(g)/(2) (2n-3) `
`=(g)/(2) (2n-4) `
or
` 40 = (10)/(2) (2n-4) ,` which gives n=3s
`therefore S= ut +(1)/(2) g t^2 `
`=0 +1/2 xx 10 xx (3)^2`
`= 45 m `