Exam

NEET

Year

2011

Subject

Physics

Chapter

Motion in Two Dimensions

NEET 2011 · Physics · Motion in Two Dimensions

A projectile is fired at an angle of 45^ with the horizontal....

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

Step-by-step solution

Height of projectile $$\begin{aligned}& H=\frac{u^2 \sin ^2 \theta}{2 g}\\ & H=\frac{u^2 \sin ^2 45^{{}^{\circ}}}{2 g}\\ & H=\frac{u^2}{4 g} \end{aligned}$$Range of projectile $$\begin{aligned}R & =\frac{u^2 \sin 2 \theta}{g}\\ & =\frac{u^2 \sin 90^{{}^{\circ}}}{g}\\ R & =\frac{u^2}{g}\\ \therefore \quad \frac{R}{2}& =\frac{u^2}{2 g}\\ \therefore \quad \tan \alpha& =\frac{H}{R / 2} \end{aligned}$$$$\begin{aligned}& =\frac{u^2 / 4 g}{u^2 / 2 g}\\ \tan \alpha& =\frac{1}{2}\\ \alpha& =\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$$