Question

A plane is inclined at an angle of $$30^{\circ} $$. A body is projected from the bottom of the inclined plane at an angle of $$60^{\circ}$$ with the ground at a speed of $$20\ m/s$$. Find the range and time of flight.

Answer 1

Correct option is B. $$ \dfrac{4}{\sqrt{3}} s , \dfrac{80}{3} m $$
$$U_{x} = 20 \cos 60 $$
$$ = 10\ m/s$$
$$U_{y} = 20 \sin 60 = 10 \sqrt{3}\ m/s$$
For time of flight
$$S_{y} = U_{y} t + \dfrac{1}{2} a_{y} t^{2} $$
$$0 = 10 \sqrt{3} \times T - \dfrac{1}{2} \times g \cos 30 \times T^{2} $$
$$T = \dfrac{2 \times 10 \sqrt{3}}{g \cos 30} = \dfrac{20 \sqrt{3}}{10 \times \sqrt{3}} \times 2 = 4 $$ sec
$$U_{x} = 20 \cos 30 = 10 \sqrt{3}\ m/s$$
$$U_{y} = 20 \sin 30 = 10\ m/s$$
$$T = \dfrac{2 U_{x}}{a} = \dfrac{2 \times 10}{g \cos 30} = \dfrac{20 \times 2}{10 \times \sqrt{3}} $$
$$ = \dfrac{4}{\sqrt{3}} $$ sec
$$R = U_{x} T + \dfrac{1}{2} a_{x} T^{2} $$
$$ = 10 \sqrt{3} \dfrac{4}{\sqrt{3}} + \dfrac{1}{2} . (-g \sin 30) \times \dfrac{16}{3} $$
$$ = 40 - \dfrac{1}{2} \times 10 \times \dfrac{1}{2} \times \dfrac{16}{3} $$
$$ = 40 - \dfrac{40}{3} = \dfrac{80}{3}\ m $$