Answer:[tex]\theta =41.409 ^{\circ}[/tex]
Explanation:
Given
Jack and Jill ran up the hill at 2.8 m/s
Horizontal component of Jill's velocity vector was 2.1 m/s
Let [tex]\theta [/tex]is the angle made by Jill's velocity with it's horizontal component
Therefore
[tex]2.8cos\theta =2.1[/tex]
[tex]cos\theta =\frac{2.1}{2.8}[/tex]
[tex]cos\theta =0.75[/tex]
[tex]\theta =41.409 ^{\circ}[/tex]
Vertical velocity is given by
[tex]V_y=2.8sin41.11=1.85 m/s[/tex]
