Answer:
Explanation:
Given
Height of counter [tex]h=1.14\ m[/tex]
Mug lands at a distance of [tex]x=1.55\ m[/tex]
Mug is released with some horizontal velocity let's say u
Consider vertical motion and using
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
y=displacement in vertical direction
u=initial velocity
a=acceleration(g)
t=time
For mug to travel 1.55 m in 0.482 s velocity is given by
[tex]u=\frac{1.55}{0.482}[/tex]
[tex]u=3.21\ m/s[/tex]
vertical velocity just before hitting
[tex]v_y=u+at[/tex]
[tex]v_y=0+9.8\times 0.482=4.72\ m/s[/tex]
Horizontal velocity is [tex]u=3.21[/tex]
[tex]\tan \theta =\frac{v_y}{u}=\frac{4.72}{3.21}[/tex]
[tex]\theta =55.77^{\circ}[/tex]
where [tex]\theta [/tex]is the angle which net velocity makes with horizontal direction in clockwise sense
