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]
