Question

If an object is thrown at an angle of ${60}^o$ with horizontal, find elevation angle of the object at its highest point as seen from the point of projection.

• A. ${\tan }^{-1}\frac{\sqrt{3}}{2}$
• B. ${\tan }^{-1}\frac{1}{2}$
• C. ${\tan }^{-1}\frac{1}{\sqrt{2}}$
• D. None of the above

Answer 1

The correct option is A ${\tan }^{-1}\frac{\sqrt{3}}{2}$

The maximum height upto which the object goes, $H=\frac{u_y^2}{2g}$ $=\frac{u^{2}si{n}^2\theta }{2g}$

The range of the motion, $R=\frac{2u_x{u}_y}{g}$ $=\frac{2u^{2}sin\theta cos\theta }{g}$

Hence the angle of elevation $=ta{n}^{-1}\left(\frac{H}{R/2}\right)$

$=ta{n}^{-1}\left(\frac{tan\theta }{2}\right)$

$=ta{n}^{-1}\left(\frac{tan{60}^{\circ }}{2}\right)$

$=ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$