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:

In a vertical projection, the loss in kinetic energy as the object rises is exactly equal to the gain in gravitational potential energy.

Answer 1

The Correct Option is C

Step 1: Use energy conservation.
Going up, mechanical energy is conserved, so kinetic plus potential energy stays constant at its initial value.
Step 2: Find the total energy.
At the launch point on the ground, $K.E. = 240$ J and $P.E. = 0$, so total energy $= 240$ J.
Step 3: Set the half-kinetic-energy target.
Half the initial kinetic energy is $\tfrac{1}{2}\times 240 = 120$ J. We want the height where $K.E. = 120$ J.
Step 4: Write the balance at that height.
$K.E. + P.E. = 240 \Rightarrow 120 + mgh = 240$, so $mgh = 120$ J.
Step 5: Substitute values.
With $m = 2$ kg and $g = 10$ ms$^{-2}$: $2\times 10\times h = 120$, i.e. $20h = 120$.
Step 6: Solve for the height.
$h = \dfrac{120}{20} = 6$ m.
\[ \boxed{6\ \text{m}} \]