vector ????⃗ has a magnitude of 17.9 and its direction is 80∘ counter‑clockwise from the x- axis. what are the x- and y- compone
1 answer:
We have vector
Therefore,
x component = 17.9 * cos80 degree = 3.108
y component = 17.9 * sin80 degrees = 17.628
<h3>What is a vector?</h3>An object with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.
If the magnitude and direction of two vectors match, they are the same vector. This shows that if we move a vector to a different location without rotating it, the final vector will be the same as the initial vector. The vectors that denote force and velocity are two examples. The direction of force and velocity are both fixed. The size of the vector would represent the force's strength or the velocity's corresponding speed.
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Answer:
The average speed of the fielder is 5.24 m/s
Explanation:
The position vector of the ball after it was hit can be calculated using the following equation:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem.
When the ball is caught, its position vector will be (see r1 in the figure):
r1 = (r1x, 0.873 m)
Then, using the equation of the position vector written above:
r1x = x0 + v0 · t · cos α
0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²
Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:
r1x = v0 · t · cos α
0.873 m = v0 · t · sin α + 1/2 · g · t²
Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:
0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²
0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²
Solving the quadratic equation:
t = 0.03 s and t = 5.72 s.
It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.
With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:
r1x = v0 · t · cos α
r1x = 37.2 m/s · 5.72 s · cos 49.3°
r1x = 1.39 × 10² m
The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.
The average velocity is calculated as the traveled distance over time, then:
average velocity = treveled distance / elapsed time
average velocity = 30.0 m / 5.72 s = 5.24 m/s
