Question

A body thrown up along a frictionless inclined plane of angle of inclination ${30}^{\circ }$ covers a distance of 40 m along the plane. If the body id projected with the same speed at an angle of ${30}^{\circ }$ with the ground, it will have a range of
$(Takeg=10m{s}^{-2})$

• A. 20 m
• B. $20\sqrt{2}m$
• C. $20\sqrt{3}m$
• D. 40 m

Answer 1

The correct option is C

$20\sqrt{3}m$

Let u be the initial speed with which the body is thrown along the inclined plane. As shown in the figure , the effective deceleration is given by
$a=gsin\theta =gsin{30}^{\circ }=\frac{g}{2}=5m{s}^{-2}$

Now, from the third equation of motion,
$v^2=u^2-2as$
$\Rightarrow 0=u^2-2\times 5\times 40$
$\Rightarrow u=20m/s$
A projectile projected at angle $\theta ={30}^{\circ }$ with this speed will have a range of
$R=\frac{u^{2}sin2\theta }{g}=\frac{20\times 20\times sin{60}^{\circ }}{10}=20\sqrt{3}m$
Hence the correct choice is (c).