Question

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is
$(g = 10\, m/s^2)$

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

Answer 1

Answer: (B)

45 m

Answer 2

To understand the question, go through the given diagram:

Fig 1: A body dropped from top of a tower fall through 40 m during the last two seconds of its fall

Let h be height of the tower and ‘t’ is the time taken by the body to reach the ground.
Here, u = 0 and a = g
$S =\, ut + \frac{1}{2}gt^2$
Where $S$ is the distance travelled by an object in time ‘t’.
Let's call height (h) as the distance travelled ($S$) by the body in time ‘t’.
$\therefore \, \, h = ut + \frac{1}{2}gt^2$ or $h = 0 \times t + \frac{1}{2}gt^2$
or, $\, \, \, h = \frac{1}{2}gt^2$ .....(i)
Distance covered in last two seconds is:
= $S - S_1 = 40$
$40 = \frac{1}{2}gt^2 - \frac{1}{2}g(t-2)^2$ (Here, $u = 0$).
or, $\, \, \, 40 = \frac{1}{2}gt^2 - \frac{1}{2}g(t^2+4-4t)$
or, $\, \, \, 40 = (2t-2)g$
or, $t = 3\, s$
From eqn (i), we get $h = \frac{1}{2} \times 10 \times (3)^2$
or, $h=45\, m$.

So, the correct option is (B): 45 m.