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 = 4.65 s and t2 = 6.24 s after it is released.

Answer 1

Answer:

84.9305655 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 4.65^2\\\Rightarrow s=106.0583625\ m$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 6.24^2\\\Rightarrow s=190.988928\ m$

The distance covered in the time interval is $190.988928-106.0583625=84.9305655\ m$