Question

A ball is projected from point $A$ with velocity $10m{s}^{-1}$ perpendicular to the inclined plane as shown in the figure. Range of the ball on the inclined plane is

• A. $\frac{40}{3}m$
• B. $\frac{20}{3}m$
• C. $\frac{12}{3}m$
• D. $\frac{60}{3}m$

Answer 1

The correct option is A $\frac{40}{3}m$

We know, for a given projectile motion, Range, $R=\frac{u^2}{g{\cos }^2\beta }[\sin (2\alpha +\beta )+\sin \beta ]$

where $u=10m/s$ initial velocity,$\alpha ={60}^{\circ }$ angle made with the horizontal
$\beta ={30}^{\circ }$ angle made with the vertical.
Substituting the values,

$R=\frac{{10}^2}{10{\cos }^2{30}^{\circ }}[\sin (2\times {60}^{\circ }+{30}^{\circ })+\sin {30}^{\circ }]$
$\Rightarrow \frac{10}{{\cos }^2{30}^{\circ }}\Rightarrow \frac{10}{\left({\frac{\sqrt{3}}{2}}^2\right)}\Rightarrow \frac{40}{3}$