A ball is thrown horizontally from the top of a tall building. At t = 2 s after projection, its velocity vector makes an angle of 45^ with the horizontal. The total straight-line displacement of the ball from its starting point at t = 3 s is : (Take g = 10 m/s ^2 and neglect air resistance)

Step-by-step solution

Let the initial horizontal velocity be u . The horizontal component of velocity remains constant, so v_x = u . The vertical component of velocity at time t is v_y = gt . At t = 2 s , the vertical velocity is: v_y = 10 2 = 20 m/s Since the velocity vector makes an angle of 45^ with the horizontal at t = 2 s : (45^ ) = v_y v_x 1 = 20 u u = 20 m/s Now, we need to find the total displacement at t = 3 s . Horizontal displacement at t = 3 s : x = ut = 20 3 = 60 m Vertical displacement at t = 3 s : y = 1 2 gt^2 = 1 2 10 (