Question

A projectile is fired with a velocity $v$ at right angle to the slope inclined at an angle $\theta$ with the horizontal. The range of the projectile along the inclined plane is

• A. $\frac{{2v}^2\tan \theta }{g}$
• B. $\frac{v^2\sec \theta }{g}$
• C. $\frac{{2v}^2\tan \theta .\sec \theta }{g}$
• D. $\frac{v^2\sin \theta }{g}$

Answer 1

Answer: (C)

$\frac{{2v}^2\tan \theta .\sec \theta }{g}$

$a_x=-g\sin \theta ;a_y=-g\cos \theta$
$u_x=0;u_y=v$
Along $y-$axis, $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$==vt-\frac{1}{2}g\cos \theta t^2$
$t=\frac{2v}{g\cos \theta }$
Along $x-$axis, $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$-R=0\times t-\frac{1}{2}g\sin \theta {\left(\frac{2v}{g\cos \theta }\right)}^2$
$R=\frac{2v^2\tan \theta \sec \theta }{g}$