A particle is projected from the ground such that its horizontal range is 4 3 times its maximum height. If the initial speed of the particle is u , the magnitude of its average velocity between the point of projection and the highest point of its trajectory is :

Step-by-step solution

Let the angle of projection be . Given the relation between horizontal range R and maximum height H : R = 4 3 H Using the standard formulas R = u^2 2 g and H = u^2 ^2 2g : u^2 (2 ) g = 4 3 ( u^2 ^2 2g ) 2 = 2 3 = 1 3 Thus, = 30^ . The average velocity vector v _ avg between the point of projection and the highest point is given by the total displacement divided by the time taken. Horizontal component of average velocity is constant and equals u . Vertical displacement is H and time taken is half the time of flight