Question

A body of mass 2 kg is thrown vertically upwards from the ground level with kinetic energy of 240 J. The kinetic energy of the body will become half at a height of (\( g=10 \text{ ms}^{-2} \)):

• A. \( 24 \text{ m} \)
• B. \( 12 \text{ m} \)
• C. \( 6 \text{ m} \)
• D. \( 4 \text{ m} \)

Hint:

The potential energy gain is equal to the loss in kinetic energy. Here, \( mgh = \Delta K.E. \).

Answer 1

The Correct Option is C

Step 1: Set up energy conservation.
On the way up, no energy is lost (we ignore air resistance), so kinetic plus potential energy stays constant.
Step 2: Find the total energy.
At the ground, $K.E. = 240$ J and $P.E. = 0$, so the total mechanical energy is $240$ J throughout the flight.
Step 3: Decide the target kinetic energy.
"Half the kinetic energy" means $K.E. = \tfrac{1}{2}\times 240 = 120$ J at the height we want.
Step 4: Balance the energy at that height.
\[ K.E. + P.E. = 240 \;\Rightarrow\; 120 + mgh = 240 \] So $mgh = 120$ J.
Step 5: Insert the numbers.
$m = 2$ kg, $g = 10$ ms$^{-2}$: $2\times 10\times h = 120$, i.e. $20h = 120$.
Step 6: Solve for height.
$h = \dfrac{120}{20} = 6$ m.
\[ \boxed{6\ \text{m}} \]