Question

A plane is inclined at an angle $\alpha ={30}^{\circ }$ with respect to the horizontal. A particle is projected with a speed $u=2\text{m/s}$ from the base of the plane, making an angle $\theta ={30}^{\circ }$ with respect to the plane as shown in the figure. The distance from the base (along the inclined plane) at which the particle hits the plane is close to

• A. $27\text{cm}$
• B. $20\text{cm}$
• C. $18\text{cm}$
• D. $14\text{cm}$

Answer 1

The correct option is A $27\text{cm}$

Given,
Speed of projection, $u=2\text{m/s}$
Inclination of plane, $\alpha ={30}^{\circ }$
Angle of projection with respect to inclined plane, $\theta ={30}^{\circ }$


We have taken x - axis along the inclined plane and y - axis perpendicular to it.
Hence,
$u_y=u\sin \theta$, $u_x=u\cos \theta$
$a_y=-g\cos \alpha$, $a_x=-g\sin \alpha$
As we know, at maximum height, velocity in $y$ direction will be zero. Hence, from $v=u+at$
$0=u\sin \theta +(-g\cos \alpha t)$
$\Rightarrow t=\frac{u\sin \theta }{g\cos \alpha }$
Time of flight $T=2t$
$T=\frac{2u\sin \theta }{g\cos \alpha }---\left(1\right)$

Distance covered in time $t$ along $x$ direction-:
$s=u_{x}T+\frac{1}{2}{a}_x{T}^2$
$x=u\cos \theta \times \frac{2u\sin \theta }{g\cos \alpha }-\frac{1}{2}\times g\sin \alpha \times \frac{4u^2{\sin }^2\theta }{g^2{\cos }^2\alpha }$
(acceleration is -ve because it is acting along -ve $x$ direction)
$\Rightarrow s=2\cos {30}^{\circ }\times \frac{2\times 2\sin {30}^{\circ }}{10\cos {30}^{\circ }}-\frac{1}{2}\times 10\sin {30}^{\circ }\times \frac{4\times 2^2\times {\sin }^{2}30}{{10}^2\times {\cos }^{2}30}$
$\Rightarrow s=0.267\text{m}=26.7$ $\text{cm}\approx 27$ $\text{cm}$
$\Rightarrow s\approx 27\text{cm}$
The distance from the base where the particle hits the plane is close to $27\text{cm}$