An ohjeet falls fhom the top of the tower travela 2,5 m in the Inst se

1.	An ohjeet falls fhom the top of the tower travela 2,5 m in the Inst second hefore sriking theground, Find theind the value of acceleration du to araviu a theeight of tho tower yVA)
Answer» It is given that distance travelled by the object 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

An ohjeet falls fhom the top of the tower travela 2,5 m in the Inst second hefore sriking theground, Find theind the value of acceleration du to araviu a theeight of tho tower yVA)

Answer»

It is given that distance travelled by the object 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

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