Question

An object, at the top of a very tall building, is released from rest and falls freely due to gravity. Neglect air resistance and calculate the distance covered by the object between times t1 = 4.12 s and t2 = 6.27 s after it is released.

Answer 1

Using 2nd kinematic equation

h = U*t + (1/2)*a*t^2

Suppose object is at height h1 after t1 time, and at height h2 after t2 time, then

distance covered between times t1 and t2 will be

d = h2 - h1

d = (U*t2 + (1/2)*a*t2^2) - (U*t1 + (1/2)*a*t1^2)

Now given that

U = initial speed = 0

t1 = 4.12 sec and t2 = 6.27 sec

a = g = 9.81 m/sec^2

So,

d = (1/2)*g*t2^2 - (1/2)*g*t1^2

d = (1/2)*g*(t2^2 - t1^2)

d = 0.5*9.81*(6.27^2 - 4.12^2)

d = 109.57 m = distance covered between t1 and t2

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