A particle is thrown with a speed v from a point O at an angle with the horizontal plane such that it passes through the point P at a height of 1 m and horizontal distance of 5 m from O, as shown in the figure. If acceleration due to gravity is g ms^ -2 , then the correct statement(s) is/are:
B. If = 45^ , the particle reaches its maximum height before it reaches P.
The equation of trajectory of a projectile is given by: y = x - g x^2 2 v^2 ^2 The particle passes through point P(5, 1). Substituting x = 5 and y = 1: 1 = 5 - 25 g 2 v^2 ^2 For option (A), substituting = 45^ : 1 = 5 45^ - 25 g 2 v^2 ^2 45^ 1 = 5(1) - 25 g 2 v^2 (1/2) 1 = 5 - 25 g v^2 25 g v^2 = 4 v^2 = 25 g 4 v = 5 g 2 ms^ -1 Thus, option (A) is correct. For option (B), the horizontal distance to the maximum height is x_m = R 2 = v^2 (2 ) 2g . For = 45^ and v^2 = 25g 4 : x_m = ( 25g 4 ) 90^ 2g = 25 8 = 3.125 m Since x_m = 3.125 m For option (C), substituting = 30^ into the trajectory equation: 1 = 5 30^ - 25 g 2 v^2 ^2 30^ 1 = 5 3 - 25 g 2 v^2 (3/4) 50 g 3 v^2 = 5 3 - 1 v^2 = 50 3 g 3(5 - 3 ) The horizontal distance to the maximum height is: x_m = v^2 60^ 2g = v^2 3 4g = 50 3 g 3(5 - 3 ) 3 4g = 25 2(5 - 3 ) 3.82 m Since x_m For option (D), substituting = ^ -1 ( 1 5 ), so = 1 5 : 1 = 5 ( 1 5 ) - 25 g 2 v^2 ^2 1 = 1 - 25 g 2 v^2 ^2 25 g 2 v^2 ^2 = 0 This is only possible if v . Thus, option (D) is incorrect. Answer: If = 45^ , then v = 5 g 2 ms^ -1 .; If = 45^ , the particle reaches its maximum height before it reaches P.
Related: Physics — Motion In Two Dimensions · All PYQ Banks