An object thrown in an upward direction does not fall the same distance in each time interval as it descends toward Earth:
because it is accelerating
Explanation:
An object thrown in an upward direction is acted upon one force only: the force of gravity, which pulls the object downward (here we are assuming air resistance is negligible).
As a result, the motion of the object is a free fall motion, which is a uniformly accelerated motion with constant acceleration [tex]g=9.8 m/s^2[/tex] downward.
Therefore, the distance covered by the object in a time t is given by the suvat equation
[tex]s=ut-\frac{1}{2}gt^2[/tex]
where u is the initial velocity.
We can check that the distance covered in one second changes as the object descends. Assuming that the initial velocity is zero, u = 0 (this assumption is irrelevant, because the term [tex]ut[/tex] is linear in t), we have:
- Displacement after 1 second:
[tex]s(1)=-\frac{1}{2}(9.8)(1)^2=-4.9 m[/tex]
- Displacement after 2 seconds:
[tex]s(2)=-\frac{1}{2}(9.8)(2)^2=-19.6 m[/tex]
So, the distance covered in the 1st second is 4.9 m, while the distance covered in the 2nd second is
[tex]|s(2)-s(1)|=|-19.6-(-4.9)|=14.7 m[/tex]
Therefore, we see that the distance covered increases every second, because the object is accelerating.
Learn more about free fall motion:
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