Question

A ball is projected from point $A$ with velocity $10\text{m/s}$ perpendicular to the inclined plane as shown in figure. Range of the ball on the inclined plane is equal to $\frac{N}{3}\text{m}$. Then, $N$ is equal to

• A. $40$
• B. $20$
• C. $12$
• D. $60$

Answer 1

The correct option is A $40$

Considering $X\&Y$ axis along the inclined plane and perpendicular to the inclined plane as shown in figure.


Along Y-axis,
$u_y=10\text{m/s}$
$a_y=-g\cos 30=\frac{-10\sqrt{3}}{2}=5\sqrt{3}{\text{m/s}}^2$
Displacement of the ball, $s_y=0$
[$\because$Ball returns to the inclined plane]
Applying kinematic equation,
$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 0=10t-\frac{1}{2}\times 5\sqrt{3}\times t^2$
$\Rightarrow t=\frac{4}{\sqrt{3}}\text{s}...\left(1\right)$
Along X-axis,
$u_x=0$
$a_x=g\sin {30}^{\circ }=10\times \frac{1}{2}=5{\text{m/s}}^2$
$s_x=$ Range
Applying kinematic equation,
$s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$s_x=\frac{1}{2}{a}_x{t}^2$
Substituting the values,
$\Rightarrow \text{Range}=\frac{1}{2}\times 5\times {\left(\frac{4}{\sqrt{3}}\right)}^2$
$\therefore \text{Range}=\frac{40}{3}\text{m}$
$\text{Range}=\frac{N}{3}\text{m}$ [According to question]
$\Rightarrow N=40$
Value of $N$ is $40$.