Question

A stone dropped from the top of the tower travels 25m in the last second of its motion. Then (g = 10 m/s square) height of the tower is

Answer 1

It is given that distance travelled by stone in last second = 25 m= Sn

let the height of tower = h

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

25 = 0 + 10/2(2n-1)

25 = 5(2n-1)

25 = 10n - 5

20= 10n

2 = n

distance travelled in last second is 2

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

25 = 10n -15

10 = 10n

n= 1

total time taken = 2 + 1 = 3 seconds

height of tower

S= ut + 1/2at²

S= 0 + 1/2 × 10 (3)²

S= 5× 9

S= 45m

Therefore, total height is 45m.