Question

A projectile is fired at an angle of $45^\circ$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

• A. $45^\circ$
• B. $60^\circ$
• C. $\tan^{ - 1} \frac{1}{2}$
• D. $\tan^{ - 1} \bigg( \frac{\sqrt 3}{2}\bigg)$

Answer 1

Answer: (C)

$\tan^{ - 1} \frac{1}{2}$

Answer 2

REF. Image. $\theta=45^{\circ}$ projection speed $=u$
horizontal velocity $ux = u c o s 45^{\circ}=\frac{ u }{\sqrt{2}}$
initial vertical velocity uy $=\operatorname{usin} 40^{\circ}=\frac{u}{\sqrt{2}}$
height $_{\max }$ attained $H =\frac{ u ^{2} \sin ^{2} \theta}{29}=\frac{ u ^{2} / 2}{29}$
$H =\frac{ u ^{2}}{49}$
half of the range $=\frac{ R }{2}=\frac{ u ^{2} \sin 2 \theta}{29}=\frac{ u ^{2} \sin 90^{\circ}}{29}$
$\frac{ R }{2}=\frac{ u ^{2}}{29}$
elevation $\phi=\tan ^{-1} \frac{ H }{ R / 2}$
$\phi=\tan ^{-1} \frac{ u ^{2} / 49}{ u ^{2} / 29}$
Or $\phi=\tan ^{-1} 0.5$