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A particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \mathrm{~ms}^{-1}$. The angle between velocity vector after $2 \mathrm{~s}$ and initial velocity vector is $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$90^{\circ}$
Initial velocity, $\mathbf{v}_{\mathbf{i}}=2 \cos \theta \hat{\mathbf{i}}+4 \sin \theta \hat{\mathbf{j}}=5 \sqrt{3} \hat{\mathbf{i}}+15 \hat{\mathbf{j}}$ Final velocity vector (after $2 s$ ),
$$
\mathbf{v}_{\mathbf{f}}=u \cos \theta \hat{\mathbf{i}}+(u \sin \theta-g t) \hat{\mathbf{j}}=5 \sqrt{3} \hat{\mathbf{i}}-5 \hat{\mathbf{j}}
$$
Now, $\mathbf{v}_{\mathrm{i}} \cdot \mathbf{v}_{\mathrm{f}}=25 \times 3-15 \times 5=0$
$$
\therefore \mathbf{v}_{\mathrm{i}} \perp \mathbf{v}_{\mathrm{f}}
$$
$$
\mathbf{v}_{\mathbf{f}}=u \cos \theta \hat{\mathbf{i}}+(u \sin \theta-g t) \hat{\mathbf{j}}=5 \sqrt{3} \hat{\mathbf{i}}-5 \hat{\mathbf{j}}
$$
Now, $\mathbf{v}_{\mathrm{i}} \cdot \mathbf{v}_{\mathrm{f}}=25 \times 3-15 \times 5=0$
$$
\therefore \mathbf{v}_{\mathrm{i}} \perp \mathbf{v}_{\mathrm{f}}
$$
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