Question

A body at rest is dropped from the top of 98 metre high tower. How much distance will travel in first second?

Answer 1

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.
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