# A plane is inclined at an angle

Question:

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

1. (1) $20 \mathrm{~cm}$

2. (2) $18 \mathrm{~cm}$

3. (3) $26 \mathrm{~cm}$

4. (4) $14 \mathrm{~cm}$

Correct Option: 1

Solution:

(1) On an inclined plane, time of flight (T) is given by

$T=\frac{2 u \sin \theta}{g \cos \alpha}$

Substituting the values, we get

$T=\frac{(2)\left(2 \sin 15^{\circ}\right)}{g \cos 30^{\circ}}=\frac{4 \sin 15^{\circ}}{10 \cos 30^{\circ}}$

Distance, $\mathrm{S}=\left(2 \cos 15^{\circ}\right) T-\frac{1}{2} g \sin 30^{\circ}(T)^{2}$

$=\left(2 \cos 15^{\circ}\right) \frac{4}{10} \frac{\sin 15^{\circ}}{10 \cos 30^{\circ}}-\left(\frac{1}{2} \times 10 \sin 30^{\circ}\right) \frac{16 \sin ^{2} 15^{\circ}}{100 \cos ^{2} 30^{\circ}}$

$=\frac{16 \sqrt{3}-16}{60} \simeq 0.1952 \mathrm{~m} \simeq 20 \mathrm{~cm}$