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A shell is fired at an angle of $30^{\circ}$ to the horizontal with velocity $196 \mathrm{~m} / \mathrm{s}$. The time of flight is $\left[\sin 30^{\circ}=\frac{1}{2}=\cos 60^{\circ}\right]$
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Verified Answer
The correct answer is:
$20 \mathrm{~s}$
We know the time of flight for a projectile is given by:
$$
t=\frac{2 u \sin \theta}{g}
$$
On placing given values in the above equation:
$$
t=\frac{20 \times(196 \mathrm{~m} / \mathrm{s}) \times \sin \left(30^{\circ}\right)}{\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)}=20 \mathrm{~s}
$$
$$
t=\frac{2 u \sin \theta}{g}
$$
On placing given values in the above equation:
$$
t=\frac{20 \times(196 \mathrm{~m} / \mathrm{s}) \times \sin \left(30^{\circ}\right)}{\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)}=20 \mathrm{~s}
$$
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