Question

A stone is allowed to fall from the top of a tower amd covers half of the height of the tower in the last second of its journey . Calculate the time taken by the stone to reach the foot of the tower.

Answer 1

Let ‘h’ be the height of the tower, ‘n’ be the time taken to reach the ground. (h,n,>0)

Using the formula S = ut + 1/2 at^2

=> h = 0 + ½ (9.8)n^2…………………….(1)

According to the problem, the stone travels h/2 in the nth second.

So, by using Sn = u + (a/2)(2n - 1) we get,

=> h/2 = 0 + (9.8/2)(2n - 1) [as initial velocity is 0]

Using (1),

[½ (9.8)n^2]/2 = (9.8/2)(2n - 1)

=> n^2/2 = 2n – 1

=> n = (2 + √2) s or (2 - √2) s
Now putting the values of n in equation (1) we get,
The time of fall is (2 + √2) s when h = 57.12 m and the time of fall is (2 - √2) s when h = 1.68 m.

Answer 2

Answer:

2+- root 2

Explanation:

Let ‘h’ be the height of the tower, ‘n’ be the time taken to reach the ground.

Using the formula S = ut + 1/2 at^2

=> h = 0 + ½ (9.8)n^2                                     --------(1)

According to the problem, the stone travels h/2 in the nth second.

So, by using Sn = u + (a/2)(2n - 1) we get,

=> h/2 = 0 + (9.8/2)(2n - 1) [as initial velocity is 0]

Using (1),

[½ (9.8)n^2]/2 = (9.8/2)(2n - 1)

=> n^2/2 = 2n – 1

=> n = (2 + √2) s or (2 - √2) s

Now putting the values of n in equation (1) we get,

The time of fall is (2 + √2) s when h = 57.12 m and the time of fall is (2 - √2) s when  h=1.68m