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:

Step-by-step solution

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 Si