Respuesta :
Answer
given,
speed of huck walk = 0.6 m/s
speed of raft on Mississippi river = 1.41 m/s
hunk velocity relative to the river bank
[tex]V_{hR} = \sqrt{V_{h}^2+V_{raft}^2}[/tex]
[tex]V_{hR} = \sqrt{0.6^2+1.41}^2}[/tex]
[tex]V_{hR} = \sqrt{2.3481}[/tex]
[tex]V_{hR} = 1.53\ m/s[/tex]
angle made with horizontal
huck move perpendicular to the shaft
[tex]\theta = tan^{-1}(\dfrac{y}{x})[/tex]
[tex]\theta = tan^{-1}(\dfrac{0.6}{1.41})[/tex]
[tex]\theta = tan^{-1}(0.426)[/tex]
θ = 23.05°
Huck's velocity relative to the river bank is [tex]1.53m/s[/tex]
The Angle from the right horizontal direction is [tex]23.05[/tex] degrees.
Resultant velocity:
Given that, Huck Finn speed is [tex]0.60 m/s[/tex] across his raft.
and the raft is traveling with speed of [tex]1.41m/s .[/tex]
So that, [tex]V_{h}=0.6m/s,V_{r}=1.41m/s[/tex]
Let us consider that Huck's velocity relative to the river bank is [tex]V_{hr}[/tex].
[tex]V_{hr}=\sqrt{V_{h}^{2}+V_{r}^{2} }\\ \\V_{hr}=\sqrt{(0.6)^{2}+(1.41)^{2} } \\\\V_{hr}=\sqrt{2.348} \\\\V_{hr}=1.53m/s[/tex]
Angle from the right horizontal direction is,
[tex]\theta=tan^{-1}(\frac{0.6}{1.41} ) \\\\\theta=tan^{-1}(0.426)\\\\\theta=23.05degrees[/tex]
Learn more about the relative velocity here:
https://brainly.com/question/17228388
