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A bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \mathrm{~km}$ away. By adjusting its angle of projection, can one hope to hit a target $5.0 \mathrm{~km}$ away?
Assume the muzzle speed to be fixed, and neglect air resistance.
Assume the muzzle speed to be fixed, and neglect air resistance.
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Verified Answer
Here $R=3 \mathrm{~km}=3000 \mathrm{~m}, \theta=30^{\circ}, g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$
As range, $R=\frac{u^2 \sin 2 \theta}{g}$
$$
\Rightarrow \quad 3000=\frac{u^2 \sin 2 \times 30^{\circ}}{9.8}=\frac{u^2 \sin 60}{9.8}
$$
$$
\Rightarrow u^2=\frac{3000 \times 9.8}{\frac{\sqrt{3}}{2}}=3464 \times 9.8
$$
Also, $R^{\prime}=\frac{u^2 \sin 2 \theta^{\prime}}{g}$ $\Rightarrow 5000=\frac{3464 \times 9.8 \times \sin 2 \theta}{9.8}$
i.e. $\sin 2 \theta^{\prime}=\frac{5000}{3464}=1.44$
which is impossible because sine of an angle cannot be more than 1. Thus this target cannot be hoped to be hit.
As range, $R=\frac{u^2 \sin 2 \theta}{g}$
$$
\Rightarrow \quad 3000=\frac{u^2 \sin 2 \times 30^{\circ}}{9.8}=\frac{u^2 \sin 60}{9.8}
$$
$$
\Rightarrow u^2=\frac{3000 \times 9.8}{\frac{\sqrt{3}}{2}}=3464 \times 9.8
$$
Also, $R^{\prime}=\frac{u^2 \sin 2 \theta^{\prime}}{g}$ $\Rightarrow 5000=\frac{3464 \times 9.8 \times \sin 2 \theta}{9.8}$
i.e. $\sin 2 \theta^{\prime}=\frac{5000}{3464}=1.44$
which is impossible because sine of an angle cannot be more than 1. Thus this target cannot be hoped to be hit.
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