51. For a given velocity of project...
Question

# 51. For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is (A) 30° (B) 45° (C) 60° (D) 90°

JEE/Engineering Exams
Physics
Solution
144
4.0 (1 ratings)
Just do it as the qquestion says. Apply the fomulae simple for up the inclined Range and down the incline Range. Therfore, after putting evrythng would cut out and the things would remain alive would be: as follows: ( 1 /(1+sin B)=3 /(1-sin B) ) where ( mathrm{B} ) is the angle of inclination of incline. Solving further, ( 1-sin B=3+3 sin B ) ( -2=4 sin B ) ( sin B=-1 / 2 ) ( mathrm{B}=120 ) degree since it is not possibe therfore the answer would be ( 180-120=60 ) degree.