Question

A stone is dropped from the top of a tower and travels 24.5 m in the last second of its journey. What is the height of its tower ?

Answer 1

Height of the tower = H
Total time taken to reach the ground = T

H = 0.5gT^2 ——(1)

H - 24.5 = 0.5g * (T - 1)^2
0.5gT^2 - 24.5 = 0.5g * (T^2 + 1 - 2T)
0.5gT^2 - 24.5 = 0.5gT^2 + 0.5g - gT
gT - 24.5 = 0.5g
10T - 24.5 = 5
10T = 29.5
T = 2.95 s

Substitute T = 2.95 s in equation (1)
H = 0.5gT^2
= 0.5 * 10 * (2.95)^2
= 43.5 m

Height of the tower is 43.5 m

[Note: Actual Height of the tower will slightly differ from 43.5 m because I took g = 10 m/s^2. If this is not your answer take g = 9.8 m/s^2 and follow the same steps.]

Answer 2

Hi Dear 
____________________________________________________________
It is given that distance travelled by stone in last second = 24.5 m 

let the height of tower = h


Sn = u + 1/2a(2n-1) 

24.5 = 0 + 10/2(2n-1)

24.5 = 5(2n-1) 

24.5 = 10n - 5 

19.5 = 10n 

1.95 = n 

distance travelled in last second is 1.95 

But distance travelled in (n-1) th second 

(Sn) = u + 1/2g(2n-1)

Sn =  5 [ 2(n-1) -1]

Sn =  5 [ 2n -2 -1 ] 

Sn = 5 [ 2n - 3 ] 

Sn = 10n - 15

Sn = 10n - 15 

24.5 = 10n -15

9.5/10 = 0.95 

total time taken = 1.95 + 0.95 = 2.9 seconds 

height of tower 

1/2gt²

5(2.9)²

5 * 8.41 = 42.5 m