Question

When jeff ran up a hill at 7.0m/s the horizontal component of velocity vector was 5.1m/s

Answer 1

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: $V=7 m/s$

Horizontal component of Jeff's velocity vector: $V_{x}=5.1 m/s$

We need to find the vertical component of Jeff's velocity vector $V_{y}$

If we observe the figure attached, both components of Jeff's velocity vector and the vector itself, form a right triangle where $V$ is the hypotenuse. Hence, we can use the pithagoream theorem to find $V_{y}$:

$V^{2}={V_{x}}^{2} +{V_{y}}^{2}$

Finding $V_{y}$:

$V_{y}=\sqrt{V^{2}-{V_{x}}^{2}$

$V_{y}=\sqrt{(7 m/s)^{2}-{(5.1 m/s)}^{2}$

Finally:

$V_{y}=4.79 m/s \approx 4.8 m/s$