Question

A stone is dropped from rest from the top of tower 19.6m high . Find the distance covered during the last second of its fall​

Answer 1

Answer:

s=14.7 m

Explanation:

We can use the formula,

[s (nth second) = u + 1/2 a (2 n - 1)],

(Where u = initial velocity

a = acceleration and s= displacement)

which can easily be derived by subtracting the displacement in (n - 1) seconds from the total displacement in 'n' seconds. (Kinematical  equation for displacement in 'n' seconds is s= un + 1/2 a n²)

Total time taken for stone to fall,

(Here acceleration = g = 9.8 m/s²)

19.6 = (0)t + 1/2(g)n²

=> n² = (19.6 x 2)/ 9.8

=> n² = 4

=> n = 2 seconds (total time taken for stone to fall)

Therefore, displacement in last, that is the 2nd second,

S = (0) + 1/2(g) (2(2) - 1)

(Here u = 0, n = 2 and acceleration = g)

=> s = 1/2 x 9.8 x 3

=> s = 14.7 m <-- Ans.