Question

A shot is fired from a point at a distance of $400m$, from the foot of a tower $100m$ high, so that it just passes over it. The direction of shot with respect to the horizontal is:

• A. $30^{\circ }$
• B. $60^{\circ }$
• C. $70^{\circ }$
• D. $45^{\circ }$

Answer 1

Answer: (D)

According to question, the given situation is shown in the figure.

Here, range, $R=200+200=400m$
Maximum height, $H=100m$
If $\theta$ be the direction of shot with respect to horizontal, then
$R=\frac{u^2\sin 2\theta }{g}$ and $H=\frac{u^2{\sin }^2\theta }{2g}$
$\therefore \frac{R}{H}=\frac{\frac{u^2\sin 2\theta }{g}}{\frac{u^2{\sin }^2\theta }{2g}}$
$\Rightarrow \frac{400}{100}=\frac{2\sin 2\theta }{{\sin }^2\theta }$
$\Rightarrow 4=\frac{2\times 2\sin \theta \cos \theta }{{\sin }^2\theta }[\because \sin 2\theta =2\sin \theta \cos \theta ]$
$\Rightarrow 1=\tan \theta$
$\Rightarrow \tan 45^{\circ }=\tan \theta$
$\Rightarrow 45^{\circ }=\theta$
$\Rightarrow \theta =45^{\circ }$