Answer:
Explanation:
Given
mass of drop [tex]m=2 kg[/tex]
height of fall [tex]h=1 m[/tex]
ball leaves the foot with a speed of 18 m/s at an angle of [tex]55^{\circ}[/tex]
Velocity of ball just before the collision with the floor
[tex]u^=2gh[/tex]
[tex]u=\sqrt{2gh}[/tex]
[tex]u=\sqrt{2\times 9.8\times 1}=4.42 m/s[/tex]
Impulse delivered in Y direction
[tex]J_y=m(v\sin (55)-(-u))[/tex]
[tex]J_y=2(18\sin (55)+4.42)[/tex]
[tex]J_y=38.32 kg-m/s[/tex]
Impulse in x direction
[tex]J_x=m\times v\cos (55)[/tex]
[tex]J_x=2\times 4.42\cos (55)=20.646[/tex]
[tex]J_{net}=\sqrt{J_x^2+J_y^2}[/tex]
[tex]J_{net}=\sqrt{(38.32)^2+(20.64)^2}[/tex]
[tex]J_{net}=43.52 kg-m/s[/tex]
at an angle of [tex]\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}[/tex]
[tex]\phi =tan^{-1}(1.856)[/tex]
[tex]\phi =61.7^{\circ}[/tex]
