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A shot is fired from a point at a distance of \(200 \mathrm{~m}\), from the foot of a tower \(100 \mathrm{~m}\) high, so that it just passes over it. The direction of shot with respect to the horizontal is
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Verified Answer
The correct answer is:
\(45^{\circ}\)
According to question, the given situation is shown in the figure.
Here, range, \(R=200+200=400 \mathrm{~m}\)
Maximum height, \(H=100 \mathrm{~m}\)
If \(\theta\) be the direction of shot with respect to horizontal, then
\(\begin{aligned}
& R=\frac{u^2 \sin 2 \theta}{g} \text { and } H=\frac{u^2 \sin ^2 \theta}{2 g} \\
& \therefore \quad \frac{R}{H}=\frac{\frac{u^2 \sin 2 \theta}{g}}{\frac{u^2 \sin ^2 \theta}{2 g}} \\
& \Rightarrow \quad \frac{400}{100}=\frac{2 \sin 2 \theta}{\sin ^2 \theta} \\
& \Rightarrow \quad 4=\frac{2 \times 2 \sin \theta \cos \theta}{\sin ^2 \theta} \quad [\because \sin 2 \theta=2 \sin \theta \cos \theta] \\
& \Rightarrow \quad 1=\tan \theta \\
& \Rightarrow \quad \tan 45^{\circ}=\tan \theta \\
& \Rightarrow \quad 45^{\circ}=\theta \\
& \Rightarrow \quad \theta=45^{\circ} \\
\end{aligned}\)
Here, range, \(R=200+200=400 \mathrm{~m}\)
Maximum height, \(H=100 \mathrm{~m}\)
If \(\theta\) be the direction of shot with respect to horizontal, then
\(\begin{aligned}
& R=\frac{u^2 \sin 2 \theta}{g} \text { and } H=\frac{u^2 \sin ^2 \theta}{2 g} \\
& \therefore \quad \frac{R}{H}=\frac{\frac{u^2 \sin 2 \theta}{g}}{\frac{u^2 \sin ^2 \theta}{2 g}} \\
& \Rightarrow \quad \frac{400}{100}=\frac{2 \sin 2 \theta}{\sin ^2 \theta} \\
& \Rightarrow \quad 4=\frac{2 \times 2 \sin \theta \cos \theta}{\sin ^2 \theta} \quad [\because \sin 2 \theta=2 \sin \theta \cos \theta] \\
& \Rightarrow \quad 1=\tan \theta \\
& \Rightarrow \quad \tan 45^{\circ}=\tan \theta \\
& \Rightarrow \quad 45^{\circ}=\theta \\
& \Rightarrow \quad \theta=45^{\circ} \\
\end{aligned}\)
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