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A projectile is projected with the velocity of $(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$. The horizontal range of the projectile will be
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Verified Answer
The correct answer is:
$2.4 \mathrm{~m}$
$2.4 \mathrm{~m}$
Given, $\mathbf{v}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
$$
\therefore \quad v=|\mathrm{v}|=\sqrt{3^2+4^2} \Rightarrow v=5 \mathrm{~m} / \mathrm{s}
$$
From figure,
$\sin \theta=\frac{4}{5}$ and $\cos \theta=\frac{3}{5}$
$\because \quad R=\frac{v^2 \sin 2 \theta}{g}$
$=\frac{v^2 \cdot 2 \sin \theta \cdot \cos \theta}{g}$
$=\frac{5 \times 5 \times 2 \times \frac{4}{5} \times \frac{3}{5}}{10}=24 \mathrm{~m}$
$$
\therefore \quad v=|\mathrm{v}|=\sqrt{3^2+4^2} \Rightarrow v=5 \mathrm{~m} / \mathrm{s}
$$
From figure,
$\sin \theta=\frac{4}{5}$ and $\cos \theta=\frac{3}{5}$
$\because \quad R=\frac{v^2 \sin 2 \theta}{g}$
$=\frac{v^2 \cdot 2 \sin \theta \cdot \cos \theta}{g}$
$=\frac{5 \times 5 \times 2 \times \frac{4}{5} \times \frac{3}{5}}{10}=24 \mathrm{~m}$
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