Question

A body of mass 2 kg is thrown up vertically with a kinetic energy of 490 J. If the acceleration due to gravity is $9.8\,m{s^{ - 2}}$, the height at which the kinetic energy of the body becomes half of the original value is:
A. 50 m
B. 25 m
C. 12.5 m
D. 10 m

Answer 1

Use law of conservation of total energy at the surface and the position where kinetic energy is half of its initial value. Assume the initial height of the ball is zero.

Formula used:
The gravitational potential energy of the body of mass m is,
$P.E = mgh$
Here, g is the acceleration due to gravity and h is the height from the surface.

Complete step by step answer:
We have given that the mass of the body is 2 kg and its initial kinetic energy is 490 J.

When we throw the body in the upward direction, the velocity of the body decreases as it moves upward. Since the kinetic energy of the body is proportional to the square of the velocity, the kinetic energy of the body also decreases while moving upward. Now, we know that the gravitational potential energy of the body is proportional to the distance from the ground. Therefore, the gravitational potential energy increases as the body moves upward.

Now, we can use the law of conservation of energy at the ground and at a distance where the kinetic energy of the body is half of its initial value.
${\left( {K.E} \right)_i} + {\left( {P.E} \right)_i} = {\left( {K.E} \right)_f} + {\left( {P.E} \right)_f}$
$\Rightarrow {\left( {K.E} \right)_i} + mg{h_i} = {\left( {K.E} \right)_f} + mg{h_f}$

Here, m is the mass of the body, g is the acceleration due to gravity, ${h_i}$ is the initial height and ${h_f}$ is the final height.

Since we have given that, the kinetic energy is half of its initial value at height ${h_f}$, we can substitute $\dfrac{{{{\left( {K.E} \right)}_i}}}{2}$ for ${\left( {K.E} \right)_f}$, 2 kg for m, 0 m for ${h_i}$ and $9.8\,m{s^{ - 2}}$ for g in the above equation.
$490 + \left( 2 \right)\left( {9.8} \right)\left( 0 \right) = \dfrac{{490}}{2} + \left( 2 \right)\left( {9.8} \right){h_f}$
$\Rightarrow 490 = \dfrac{{490}}{2} + 19.6{h_f}$
$\Rightarrow {h_f} = \dfrac{{\left( {490 - \dfrac{{490}}{2}} \right)}}{{19.6}}$
$\Rightarrow {h_f} = 12.5\,m$

So, the correct answer is “Option C”.

Note:
To solve such types of questions, you can blindly use the law of conservation of energy if there is no loss in the total energy. In this question, we have assumed that the initial height of the ball is zero and it is thrown from the ground. But in some of the questions, the initial height of the ball is given which will be the height of the person throwing the ball.