Question

A body initially at rest at a certain height from the ground is falling freely under gravity. At a time of one second, the body is at a height of 120 m from the ground. At a time $t=4.5\,s$, if the potential energy of the body is K times its total energy, then the value of K is (Acceleration due to gravity $=10\,ms^{-2}$)

• A. 0.19
• B. 0.81
• C. 0.37
• D. 0.63

Hint:

For freely falling bodies: \[ \frac{PE}{\text{Total Energy}} = \frac{\text{Current Height}}{\text{Initial Height}} \]

Answer 1

The Correct Option is A

Concept: During free fall, the total mechanical energy remains constant. \[ E=PE+KE \] The ratio \[ K=\frac{PE}{E} \] can be determined from the height of the body at the specified instant.

Step 1: Find the initial height.
At \(t=1\,s\), \[ s=\frac12 gt^2 \] \[ s=\frac12(10)(1)^2 \] \[ s=5\,m \] Height after one second is \(120\,m\). Therefore, \[ H-5=120 \] \[ H=125\,m \]

Step 2: Find height at \(t=4.5\,s\).
Distance fallen: \[ s=\frac12(10)(4.5)^2 \] \[ s=5(20.25) \] \[ s=101.25\,m \] Hence remaining height, \[ h=125-101.25 \] \[ h=23.75\,m \]

Step 3: Calculate the ratio of potential energy to total energy.
Total energy \[ E=mgH \] Potential energy at \(t=4.5\,s\), \[ PE=mgh \] Therefore, \[ K=\frac{PE}{E} \] \[ K=\frac{h}{H} \] \[ K=\frac{23.75}{125} \] \[ K=0.19 \] Hence, \[ \boxed{K=0.19} \]