If we throw a body upwards with velocity of 4 ms⁻¹ at what height its kinetic energy reduces to half of the initial value ? Take g = 10 m /s²

The correct option is None of these.

To find the height at which the kinetic energy of the body reduces to half of its initial value, we can use the law of conservation of mechanical energy.

Let the mass of the body be $m$.
The initial velocity with which the body is thrown upwards is:
$u=4{\text{ms}}^{-1}$
The acceleration due to gravity is:
$g=10{\text{m/s}}^2$

The initial kinetic energy ($K_i$) of the body at the ground level (height = 0) is given by:

$K_i=\frac{1}{2}m{u}^2$

At a certain height $h$, the kinetic energy reduces to half of its initial value. Therefore, the final kinetic energy ($K_f$) at height $h$ is:

$K_f=\frac{1}{2}$

According to the law of conservation of mechanical energy, the decrease in kinetic energy is equal to the increase in potential energy ($U=mgh$):

$K_i-K_f=mgh$

Substituting the values of $K_i$ and $K_f$:

$K_i-\frac{1}{2}$

$\frac{1}{2}$

Substitute the expression for $K_i$ into the equation:

$\frac{1}{2}$

$\frac{1}{4}m{u}^2=mgh$

We can cancel out the mass $m$ from both sides:

$h=\frac{u^2}{4g}$

Now, substitute the given numerical values of $u$ and $g$:

$h=\frac{4^2}{4\times 10}$

$h=\frac{16}{40}=0.4\text{m}$

Since $0.4\text{m}$ is not present in options (4m, 2m, or 1m), the correct choice is None of these.