Q. A body is projected with initial velocity of $(8\widehat{\mathbf{i}} + 6\widehat{\mathbf{j}})ms^{- 1}$. The horizontal range is
A) 9.6 m
B) 14 m
C) 50 m
D) None of these
Explanation
(a) Here,
$v = \sqrt{(8)^{2} + (6)^{2}} = 10$ and $\tan\theta = \frac{6}{8}$
$\therefore\ $Hypotenuse, $h = 10$m
$\therefore\ sin\ \theta = \frac{6}{10},\cos{\theta = \frac{8}{10}}$
$R = \frac{2u^{2}\sin{\theta\cos\theta}}{g}$
$R = \frac{2 \times 10 \times 10 \times \frac{6}{10} \times \frac{8}{10}}{10}$
$R = \frac{96}{10} = 9.6m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.