Question

A plane is inclined at an angle $\alpha ={30}^o$ with a respect to the horizontal. A particle is projected with a speed $u=2m{s}^{-1}$ from the base of the plane, making an angle $\theta ={15}^o$ with respect to the plane as shown in the figure.The distance from the base, at which the particle hits the plane is close to :
(Take $g=10m{s}^{-2}$)

• A. $14cm$
• B. $20cm$
• C. $18cm$
• D. $26cm$

Answer 1

The correct option is B $20cm$

$t=\frac{2\times 2\times \sin {15}^o}{g\cos {30}^o}$
$S=2\cos {15}^o\times t-\frac{1}{2}g\sin {30}^o{t}^2$
put values and solve
$S\simeq 20cm$