A body at rest is dropped from the top of 98 metre high tower. How much distance will travel in first second?
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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.
hope you like this
[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.
hope you like this
samtennyson473:
what?
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