Answer:
Option D) Both B and C
Explanation:
A textbook is dropped from the 2nd story stairs and falls freely. When an object falls freely, the only force acting on it is the force of gravity which results in a constant acceleration of g. So, whenever an object falls freely, its acceleration will be constant and equal to g(9.8 m/s²). Therefore, the acceleration of the textbook will remain constant. Option C is a correct option.
Since, the acceleration is constant and equal to 9.8 m/s², this implies that with every second, the velocity of the textbook will increase by 9.8 m/s. Since initial velocity is 0, the velocity after 1st second would be 9.8 m/s so that the acceleration is constant and equal to 9.8 m/s². Similar, after 1 sec the velocity is 9.8 m/s, so the velocity after 2 seconds would be 19.6 m/s so that the acceleration is constant and equal to 9.8 m/s². This way, after every second the velocity of the textbook, and every free fall object, increases by 9.8 m/s in ideal cases. So, option B is also correct.
Therefore, Both B and C are correct for the given scenario.
The textbook is in free fall motion because it is falling (in motion) due to gravity alone
The changes that will be observed are;
The velocity increases and the acceleration remains constant
The correct option is; Both B & C
Reason:
The type of motion of a text book dropped from the second story stairs is known as free fall motion
The kinematic equation of the motion is v = u + g·t
Where;
v = The velocity of the textbook
u = The initial velocity of the textbook = 0 m/s
t = The time after which it is dropped
g = The constant acceleration due to gravity ≈ 9.81 m/s²
Which gives;
v = g·t
Therefore, the velocity of the textbook increases as the time of free fall increases while the acceleration, g remains constant at 9.81 m/s²
Therefore, the correct option is both B & C
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