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

Question

Aicharya · Accepted Answer

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 $\begin{aligned} & R=\frac{u^2 \sin 2 \theta}{g} \text { and } H=\frac{u^2 \sin ^2 \theta}{2 g} \\ & \therefore \quad \frac{R}{H}=\frac{\frac{u^2 \sin 2 \theta}{g}}{\frac{u^2 \sin ^2 \theta}{2 g}} \\ & \Rightarrow \quad \frac{400}{100}=\frac{2 \sin 2 \theta}{\sin ^2 \theta} \\ & \Rightarrow \quad 4=\frac{2 \times 2 \sin \theta \cos \theta}{\sin ^2 \theta} \quad [\because \sin 2 \theta=2 \sin \theta \cos \theta] \\ & \Rightarrow \quad 1=\tan \theta \\ & \Rightarrow \quad \tan 45^{\circ}=\tan \theta \\ & \Rightarrow \quad 45^{\circ}=\theta \\ & \Rightarrow \quad \theta=45^{\circ} \\ \end{aligned}$

A shot is fired from a point at a distance of \(200 \mathrm{~m}\), from the foot of a tower \(100…

Solution