If a sphere and paper are dropped from 10 meters and the sphere hit the ground in 1.4 seconds and the paper hit in 2 seconds, why did the impact time in case of the piece of paper change in comparison to the sphere?
A. a=(Wt-R)/m;a=g B. a=(Wt-R)/m;a>g
C. a=(Wt-R);a<g D. a=(Wt-R)/m;a<g
To understand why the impact time in the case of the piece of paper changed in comparison to the sphere, we need to consider the concepts of acceleration, weight, and resistance.
When an object falls, it experiences acceleration due to gravity (denoted as 'g'). The weight ('Wt') of an object is the force exerted on it due to gravity and is calculated as the mass ('m') of the object multiplied by the acceleration due to gravity ('g').
In this case, both the sphere and the paper are dropped from the same height of 10 meters. However, the time taken for the sphere to hit the ground is 1.4 seconds, while it takes 2 seconds for the paper to hit the ground.
The difference in impact time is due to the presence of air resistance acting on the paper. Air resistance is a force that opposes the motion of an object through the air. Unlike the sphere, which is relatively compact and heavy, the paper is lightweight and has a larger surface area, resulting in more air resistance acting on it.
To determine the impact time, we can use the equation:
a = (Wt - R) / m
where 'a' is the acceleration of the falling object, 'Wt' is the weight of the object, 'R' is the resistance or air resistance acting on the object, and 'm' is the mass of the object.
In the case of the paper, the air resistance (R) plays a significant role in reducing the acceleration (a) of the falling object. Therefore, the correct answer is option B:
a = (Wt - R) / m; a > g
This means that the acceleration of the paper is less than the acceleration due to gravity since the air resistance is slowing it down.