Question

A projectile is fired at angle of ${45}^{\circ }$ with the horizontal. Elevation angle of the projectile at its heighest point as seen from the point of proejection, is

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

Answer 1

The correct option is D ${\tan }^{-1}\frac{1}{2}$

REF. Image.

$\theta ={45}^{\circ }$ projection speed = u

horizontal velocity $ux=ucos{45}^{\circ }=\frac{u}{\sqrt{2}}$

initial vertical velocity uy $=usin{40}^{\circ }=\frac{u}{\sqrt{2}}$

$heigh{t}_{max}$ attained H$=\frac{u^{2}si{n}^2\theta }{29}=\frac{u^2/2}{29}$

$H=\frac{u^2}{49}$

half of the range $=\frac{R}{2}=\frac{u^{2}sin2\theta }{29}=\frac{u^{2}sin{90}^{\circ }}{29}$

$\frac{R}{2}=\frac{u^2}{29}$

elevation $\varphi =ta{n}^{-1}\frac{H}{R/2}$

$\varphi =ta{n}^{-1}\frac{u^2/49}{u^2/29}$

Or $\varphi =ta{n}^{-1}0.5$