A ball of mass 2 kg thrown from a tall building with velocity $\overrightarrow{\mathrm{v}}=(20\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+(24\mathrm{m}/\mathrm{s})\hat{\mathrm{j}}$ at time t = 0 s. Change in the potential energy (in kJ) of the ball after t = 8 s is (The ball is assumed to be in air during its motion between 0 s and 8 s, $\hat{\mathrm{i}}$ is along the horizontal and $\hat{\mathrm{j}}$ is along the vertical direction, is $\_\_\_\_\_\_$ – x × 10^–2 (g = 10m/s²) then the value of x is

Given,

mass of the ball is m = 2kg

velocity = $\overrightarrow{\mathrm{v}}=(20\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+(24\mathrm{m}/\mathrm{s})\hat{\mathrm{j}}$

$∆\mathrm{P}=∆\mathrm{KE}=\frac{1}{2}({\mathrm{u}}^2-{\mathrm{v}}^2)$

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\mathrm{v}=(20\hat{\mathrm{i}}+24\hat{\mathrm{j}})-10\hat{\mathrm{j}}\times 8$

$\mathrm{v}=20\hat{\mathrm{i}}+56\hat{\mathrm{j}}$

In accordance with the conservation of energy principle.
Potential energy change equals the change in kinetic energy.

$$\begin{gathered} \frac{1}{2}{\mathrm{mu}}^2-\frac{1}{2}{\mathrm{mv}}^2 \\ =\frac{1}{2}\mathrm{m}\left[(\mathrm{u}.\mathrm{u})-\left(\mathrm{vv}\right)\right] \\ =\frac{1}{2}\left(2\right)\left[(20\hat{\mathrm{i}}+24\hat{\mathrm{j}})(20\hat{\mathrm{i}}+24\hat{\mathrm{j}})-\left\{\right(20\hat{\mathrm{i}}-56\hat{\mathrm{j}}\left)\right(20\hat{\mathrm{i}}-56\hat{\mathrm{j}})\right] \\ =(20\times 20+24\times 24)(20\times 20+56\times 56) \\ =-2.56\mathrm{kJ} \end{gathered}$$

Hence the correct answer is 256.