Question

A projectile is fixed from level ground at an angle $\theta$ above the horizontal. The elevation angle $\varphi$ of the highest point as seen from the launch point is related to $\theta$ by the relation

• A. $tan\varphi =\frac{1}{4}tan\theta$
• B. $tan\varphi =tan\theta$
• C. $tan\varphi =\frac{1}{2}tan\theta$
• D. $tan\varphi =2tan\theta$

Answer 1

The correct option is C $tan\varphi =\frac{1}{2}tan\theta$

From figure, $tan\varphi =\frac{h}{R/2}$

We know that, range of projectile $R=\frac{u^{2}sin2\theta }{g}$

Maximum height attained by the projectile $h=\frac{u^{2}si{n}^2\theta }{2g}$

Thus we get $tan\varphi =\frac{u^{2}si{n}^2\theta /2g}{u^{2}sin2\theta /2g}=\frac{si{n}^2\theta }{sin2\theta }$

$\therefore \tan \varphi =\frac{si{n}^2\theta }{2sin\theta cos\theta }=\frac{sin\theta }{2cos\theta }=\frac{1}{2}tan\theta$