Physics Question on Motion in a plane

Question

A plane is inclined at an angle $\alpha = 30^{\circ}$ with a respect to the horizontal. A particle is projected with a speed $u = 2 ms^{-1}$ from the base of the plane, making an angle $q = 15^{\circ}$ 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 $g = 10 \; ms^{-2}$ )

Answer 1

Solution

$t = \frac{2\times2\times\sin 15^{\circ}}{g \cos 30^{\circ}}$
$S= 2 \cos 15^{\circ} \times t - \frac{1}{2} g \sin30^{\circ} t^{2}$
Put values and solve
S $\simeq$ 20cm