Question

16 A ball is kicked towards goal posts from a position 20 m from and directly in front of the posts. The ball takes 0.60 s from the time it is kicked to pass over the cross-bar, 2.5 m above the ground. The ball is at its maximum height as it passes over the cross-bar. You may ignore air resistance. a Calculate the ball's horizontal component of velocity b Calculate the vertical component of the velocity of the ball immediately after it is kicked. [2] c Determine the magnitude of the initial velocity of the ball immediately after it is kicked. 2] d Determine the angle above the horizontal at which the ball is kicked. 12]

Answer 1

Extence = 20 ml time taken = 0.60 sec Horizontal velocity U_x = 20/0.6 U_1 = 33.34 mol^-1 For Finding vertical velocity U_y, 0 We have time taken = 0.60 sec Height h = 2.5 m aceelestation g = 9.8 ms^-2 using S = u + 1/2 gt^2 Put s = distance covered = 2.5 m t = 0.60 sec g = -9.8 ms^-2 (direction downwards) 2.5 = u_y(0.6) -1/2 times 9.8 times (0.6) ^2 U_y0 = 7.11 ms^-1 (initial vertical velocity) Magnitude of initial velocity = squareroot (U_1) ^2 + (U_y) ^2 = squareroot (33.34) ^2 + (7.11) ^2 = 34.09 ms^-1 Angle From Horizontal = Tan^-1(U_y/U_x) = 12.03 degree