Question

A body is dropped form the top of a tower. It falls through 40 m during the last two seconds of its fall.The height of tower in m is $(g=10m{s}^{-2})$

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

Answer 1

The correct option is A 45 m

In the case of free fall displacement of the motion is
$S=\frac{1}{2}g{t}^2$

let assume

nth second = $t_n$

(n-2)th second = $t_{n-2}$

given that ${(t}_n-t_{n-2})=2...(1)$

Displacement in last two seconds,

$S_{{(n,n-1)}^{th}}=\frac{1}{2}g({t_n{-t}_{n-2})}^2$

putting values in this equation

$⟹40=\frac{1}{2}\times 10\times ({t_n{-t}_{n-2})}^2$

$⟹\left(t_n{-t}_{n-2})\right(t_n{+t}_{n-2})=8$

$⟹(t_n{+t}_{n-2})=4...(2)$

by solving eq 1 and eq 2 we get the values of

$⟹t_n=3⟹t_{n-2}=1$

hight of the building

$S=\frac{1}{2}g{t}^2$

$⟹S=\frac{1}{2}\times 10\times 3^2$

$⟹S=45m$