Answer: 4.8 m/s
Explanation:
The complete question is written below:
Jeff ran up the hill at 7.0 m/s. The horizontal component of Jeff's velocity vector was 5.1 m/s. What was the vertical component of Jeff's velocity?
We have the following data:
Jeff's velocity: [tex]V=7 m/s[/tex]
Horizontal component of Jeff's velocity vector: [tex]V_{x}=5.1 m/s[/tex]
We need to find the vertical component of Jeff's velocity vector [tex]V_{y}[/tex]
If we observe the figure attached, both components of Jeff's velocity vector and the vector itself, form a right triangle where [tex]V[/tex] is the hypotenuse. Hence, we can use the pithagoream theorem to find [tex]V_{y}[/tex]:
[tex]V^{2}={V_{x}}^{2} +{V_{y}}^{2}[/tex]
Finding [tex]V_{y}[/tex]:
[tex]V_{y}=\sqrt{V^{2}-{V_{x}}^{2}[/tex]
[tex]V_{y}=\sqrt{(7 m/s)^{2}-{(5.1 m/s)}^{2}[/tex]
Finally:
[tex]V_{y}=4.79 m/s \approx 4.8 m/s[/tex]
