A ball of mass $m=1 \mathrm{~kg}$ is thrown from the top of a building with initial velocity $\mathbf{v}=(20 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(24 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}$ at time $t=0$. The change in the potential energy of the ball between $t=0$ and $t=6 \mathrm{~s}$, if the ball does not hit the ground, then (assume, $g=10 \mathrm{~m} \mathrm{~s}^2$ )

• A

  $-320 \mathrm{~J}$
• Correct answer
  B

  $-360 \mathrm{~J}$
• C

  $-380 \mathrm{~J}$
• D

  $320 \mathrm{~J}$

Solution

$\mathbf{u}=20 \hat{\mathbf{i}}+24 \hat{\mathbf{j}}$
To find change in PE, we have to consider only vertical motion of the ball. So, we have $u_y=24 \mathrm{~m} / \mathrm{s}, m=1 \mathrm{~kg}$ $a_y=-10 \mathrm{~m} / \mathrm{s}^2, t=6 \mathrm{~s}$ We take origin at point of projection, $h=u_y t-\frac{1}{2} g t^2=24 \times 6-\frac{1}{2} \times 10 \times 36$ $=144-180=-36 \mathrm{~m}$ So, vertical displacement of the particle is $h=-36 \mathrm{~m}$ Hence, change in PE $=m g h=1 \times 10 \times-36=-360 \mathrm{~J}$

Hence, the correct answer is option: 'B'.