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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, $g=10 \mathrm{~m} \mathrm{~s}^2$ )
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Verified Answer
The correct answer is:
$-360 \mathrm{~J}$
$\mathbf{u}=20 \hat{\mathbf{i}}+24 \hat{\mathbf{j}}$
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 $=m g h=1 \times 10 \times-36=-360 \mathrm{~J}$
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 $=m g h=1 \times 10 \times-36=-360 \mathrm{~J}$
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