Question

A stone is dropped from a height of 45m. What will be the distance travelled by it during the last second of its motion?(Take $g=10m{s}^{-2}$)

• A. 30 m
• B. 25 m
• C. 15 m
• D. 5 m

Answer 1

Answer: (B)

25 m

When the stone is dropped, initial velocity = 0.
Now, $h=\left(1/2\right)g{t}^2$
Distance travelled in $t_n$ sec can be written as $h_n=\left(1/2\right)g{t_n}^2$
Distance travelled in $t_n-1$ sec can be written as $h_{n-1}=\left(1/2\right)g(t_n-1{)}^2$
So, $h_n-h_{n-1}$ = journey in nth second = $\left(1/2\right)g(t_n^2-{(t_n-1)}^2)$
$=\left(1/2\right)g(2t_n-1)=g(t_n-1/2)$
Now using $h=\left(1/2\right)g{t}^2$ for h = 45m we get t = 3 sec
So, in the 3rd second the stone will travel 10 X 2.5 m = 25 m.