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 for the paper changed compared to the sphere, we need to consider the forces acting on the two objects during their free fall.
Let's analyze the situation using the equations of motion:
1. The equation for distance traveled during free fall is given by: d = 0.5 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2 on Earth), and t is the time.
2. The equation for final velocity is given by: v = g * t, where v is the final velocity.
Now, in the case of the sphere, let's assume it experiences negligible air resistance. Therefore, the only significant force acting on it is the force of gravity. Hence, we can use the equation for distance to determine the time taken for it to hit the ground:
10 = 0.5 * 9.8 * t^2
t^2 = 10 / 4.9
t^2 = 2.04
t ≈ 1.43 seconds
Now, let's consider the piece of paper. Unlike the sphere, the paper is affected by air resistance, which creates an additional force opposing its motion. Because of this opposing force, the paper slows down more quickly than the sphere, resulting in a longer time for it to reach the ground. Therefore, the impact time for the paper is greater than that of the sphere.
Based on this analysis, the correct answer is option D. a=(Wt-R)/m;a<g, which indicates that the acceleration of the paper (a) is equal to the net force (weight minus air resistance) divided by its mass, and the acceleration is less than the acceleration due to gravity (g).