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# 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.

A
$$4s , \dfrac{80}{3}\ m$$
B
$$4s , \dfrac{75}{3} m$$
C
$$\dfrac{4}{\sqrt{3}} s , \dfrac{75}{3}\ m$$
D
$$\dfrac{4}{\sqrt{3}} s , \dfrac{80}{3} m$$
Solution
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#### 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 $$R = U_{x} T + \dfrac{1}{2} a_{x} T^{2}$$ $$= 10 \sqrt{3} \dfrac{4}{\sqrt{3}} + \dfrac{1}{2} . (-g \sin 30) \times \dfrac{16}{3}$$ $$= 40 - \dfrac{1}{2} \times 10 \times \dfrac{1}{2} \times \dfrac{16}{3}$$ $$= 40 - \dfrac{40}{3} = \dfrac{80}{3}\ m$$

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