What is the distance covered by a freely falling body during the first three seconds of itsmotion? (g=10m/s2
Question
What is the distance covered by a freely falling body during the first three seconds of its motion?
(g = 10 m/s²)
Solution
The distance covered by a freely falling body can be calculated using the equation of motion:
s = ut + 0.5gt^2
where: s = distance covered u = initial velocity t = time g = acceleration due to gravity
In this case, the body is freely falling, so the initial velocity (u) is 0. The acceleration due to gravity (g) is given as 10 m/s^2, and the time (t) is 3 seconds.
Substituting these values into the equation gives:
s = 03 + 0.5103^2 s = 0 + 0.510*9 s = 0 + 45 s = 45 meters
So, the distance covered by the body during the first three seconds of its motion is 45 meters.
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