Since, the stone is dropped, its initial horizontal velocity is equal to the velocity og the bus which is, ux = 60 km/h = 16.67 m/s Its initial vertical velocity is, u y = 0 The height through which the stone falls is, h = 1.96 m Using, S = ut + ½ at 2 => 1.96 = 0 + ½ (9.8)(t 2 ) => t = 0.63 s Horizontal distance moved by the stone in this time is, X = u x t = (16.67)(0.63) = 10.5 m The path of the stone can be found as follows: X = u x t = 16.67t Y = u y t + ½ at 2 = (1/2)(9.8)t 2 = 4.9t 2