A particle is projected from ground At a height of 0.4 m from the ground, the velocity of a projectile in vector form is v = (6i + 2j) m/s (the x-axis is horizontal and y-axis is vertically upwards).

Question

A particle is projected from ground At a height of 0.4 m from the ground, the velocity of a projectile in vector form is v = (6i + 2j) m/s (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is (g = 10 m/s²):

Prashant Kotian · Accepted Answer

Correct answer: C. Let u be the initial speed of the particle Then v2 = u2 – 2gh or u2 = v2 + 2gh or 2 2 x y u u  = 2 2 x y v v 2gh   (vx = ux) or 2 y u = 2 y v 2gh  or 2 y u = (2)2 + (2) (10) (0.4) = 12  uy = 12 or 2 3 m / s and ux = vy = 6 m/s  tan  = y x u u = 2 3 1 6 3  or  = 30°

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