Question

A stone is dropped from the top of a tower travels 25 m in the last second of it's motion of g=10ms^2 height of the tower is?​

Answer 1

Answer:

45m

Explanation:

Distance travelled in the last second of its motion is 25m

We know that,

'S' in nth second of its motion = u + a/2 (2n-1)

So in this problem ,

S in nth second = 0 + g/2(2n-1)

25 = g/2 (2n-1)

5 = 2n - 1

n=3 So its total time of descent is 3 seconds

For a freely falling body,

h = 1/2 gt²

  = 1/2 × 10 × 9

  = 45 m   ⇒ this is the height of the tower