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

The Correct Option is B

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.