A body dropped from top of tower falls through 40m . . .

Question : A body dropped from top of tower falls through 40m during last two seconds of its fall . Calculate height of tower? (Take g=10 m/s²)

Doubt by Zoha 

Solution : 

Let AC = h = height of the building. 

Let B is the point such that BC = 40 m
Time taken to cover the distance BC = t'= 2 sec. 

Let the velocity of the body at point B be u'

Using 
s = ut+½at²

BC = u't'+½gt'²
40 = u'(2)+½(10)(2)²
40 = 2u' + (40/2)
40 = 2u' + 20
40-20 = 2u'
20/2 = u'
u' = 10 m/s

Now, 

Let v be the final velocity of the body before touching the ground at point C. 

Now once again considering the motion of the body from B to C

Using Equation 

v=u+gt
v=u'+gt'

v=10+(10)2
v = 10+20
v = 30 m/s

Now Considering the motion of the body from A to C

Initial velocity (u) = 0 m/s

Final velocity before touching the ground (v) = 30 m/s

Using 
v²-u² = 2as

v²-u² = 2gh

(30)²-(0)² = 2(10)h
900 - 0 = 20h
900/20 = h

90/2 = h
45 = h
h = 45 m

Hence, the required height of the tower is 45 m.

Alternate Method 

Consider the motion of the body from A to C 

Here, 

u = 0 m/s
s = AC = h m

time = t sec

Using 

s=ut+½at²

s=ut+½gt²

h=(0)t + ½gt²
h=½gt² — (1)

Now, considering the motion of the body from A to B

Here 

u = 0 m/s
s = AB = h₁ m

time (t') = (t-2) sec

Using 

s=ut+½at²

s=ut'+½gt'²
h₁=(0)(t-2)+½g(t-2)²
h₁=0+½g(t-2)²
h₁=½g(t-2)²

Now 

BC = 40 m (Given)
AC-AB = 40

h-h₁=40

½gt²-½g(t-2)² = 40
½g[t²-(t-2)²] = 40
[t²-(t-2)²] = (40×2)g

t²-(t²+4-4t) = 80/g [∵g=10 m/s²]
t²-t²-4+4t = 80/10
-4+4t = 8
4t = 8+4
4t = 12
t = 12/4
t = 3 sec

putting in equation (1)

h=½g(3)²
h=½(10)(9)
h = 90/2
h = 45 m
Hence, the required height of the tower is 45 m.