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Question: Consider the following statements regarding the K.E of a freely falling body: a) Is directly propo...

Consider the following statements regarding the K.E of a freely falling body:
a) Is directly proportional to height of its fall
b) Is inversely proportional to height of its fall
c) Is directly proportional to square of time of its fall
d) Is inversely proportional to square of time of its fall

A. a, b & d are correct
B. b, c & d are correct
C. a & c are correct
D. a, b, c & d are correct

Explanation

Solution

Recall the formula for the kinetic energy of the body in terms of velocity of the body. Use the kinematic equations in the kinetic energy formula to check the dependence of kinetic energy on the height of fall and time. The initial velocity of the body falling freely is always equal to zero.

Formula used:
Kinetic energy, K=12mv2K = \dfrac{1}{2}m{v^2}
Here, m is the mass of the body and v is the velocity.
v=u+gtv = u + gt, where, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and t is the time.

Complete step by step answer:
We know that for the freely falling body, its initial velocity is zero. We can determine the final velocity, distance travelled by the body in time t using kinematic equations. We know that for a freely falling body, the acceleration in the body is the acceleration due to gravity. We have the expression for kinetic energy of the body moving with velocity v,
K=12mv2K = \dfrac{1}{2}m{v^2} …… (1)
Here, m is the mass of the body.

We have the kinematic equation,
v=u+gtv = u + gt
Since the initial velocity of the body is zero, we can write the above equation as,
v=gtv = gt
Substituting v=gtv = gtin equation (1), we get,
K=12m(gt)2K = \dfrac{1}{2}m{\left( {gt} \right)^2}
K=12mg2t2\Rightarrow K = \dfrac{1}{2}m{g^2}{t^2}

Since the mass and acceleration due to gravity are constants, we can say that the kinetic energy of the body is proportional to the square of the time.

We also have the kinematic equation,
v2=u2+2hs{v^2} = {u^2} + 2hs

Since the initial velocity of the body is zero, we can write the above equation as,
v2=2gh{v^2} = 2gh
Substituting the above equation in equation (1), we get,
K=12m(2gh)K = \dfrac{1}{2}m\left( {2gh} \right)
K=mgh\Rightarrow K = mgh
Thus, from the above equation, we can say that the kinetic energy of the body is proportional to the height of the body above the ground. Therefore, the correct statements are a) and c).

So, the correct answer is option C.

Note: For the vertical motion of the body, we have replaced acceleration a with the acceleration due to gravity g in the kinematic equation. Also, the vertical distance is taken as height of the body. For the vertical downward motion of the body, always take the acceleration due to gravity and height as positive value.