Question

A stone is dropped from a height of 19.6m. What distance it travels during (1) the first 0.1sec,
(ii) the last 0.1sec of its motion? Take ga 9.8m/s2
(A: 0.049m, 1.9m]​

Answer 1

Explanation:

We can use the formula,

[s (nth second) = u + 1/2a (2n - 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 + 1/2an²)

Total time taken for stone to fall,

(Here acceleration = g = 9.8m/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 = 14.7m <-- Ans.

Answer 2

Answer:

sorry I don't know

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