Answer
given,
height of counter = 1.20 m
horizontal distance from base = 1.6 m
a) velocity of mug = ?
  using equation of motion
  [tex]s_y = u t + \dfrac{1}{2}gt^2[/tex]
  [tex]1.2 = \dfrac{1}{2}\times 9.8 \times t^2[/tex]
  [tex]t^2 = 0.245[/tex]
   t = 0.495 s
 speed of the mug
   s_x = v x t
   1.6 = v x 0.495
   v = 3.23 m/s
b) final velocity of mug in y direction
   again using equation of motion
   v² = u² + 2 a s
   v²= 0 + 2 x 9.8 x 1.2
   v = √23.52
   v_y = 4.85 m/s
now, direction
 [tex]\theta = tan^{-1}(\dfrac{v_y}{v_x})[/tex]
 [tex]\theta = tan^{-1}(\dfrac{4.85}{3.23})[/tex]
 [tex]\theta =56.34^0[/tex]
