X and Y components of a vector using a graph of speed/time

In summary, the golf ball is struck at ground level with a speed of 29.3 m/s. The graph shows the speed as a function of time, with 16 m/s as the minimum value and 29.3 m/s as the maximum value. To find the horizontal distance traveled before returning to ground level, we can use the formula s=.5(v0+v)t, which gives us a distance of 80m. To find the maximum height above ground level attained by the ball, we can use the formula s=.5(v0+v)t, which gives us a height of 30.6825m. The initial velocity of the ball is a mixture of both x and y components, with a magnitude of
  • #1
AidenPhysica
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Homework Statement



A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in the figure, where t = 0 at the instant the ball is struck. The scaling on the vertical axis is set by va= 16 m/s and vb= 29.3 m/s. (a) How far does the golf ball travel horizontally before returning to ground level? (b) What is the maximum height above ground level attained by the ball?
[PLAIN]https://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c04/fig04_40.gif[/B][/SIZE]

Homework Equations


x component=|magnitude of vector|cos(theta)
y component=|magnitude of vector|sin)theta)
s=.5(v0+v)t

The Attempt at a Solution


16 m/s is the minimum value on this graph and 29.3 is the maximum value on this graph. 29.3 appears first and then 16 is in the middle in a concave-up parabola. So this is what I did and I wonder if this is right:
So the initial velocity, a mixture of both x and y components is=29.3. To find the theta, we can take 29.3cos(theta)=16. Why do we use 16? Because the minimum speed corresponds to a vertical velocity of 0, meaning 16 is the constant horizontal velocity. So then, theta is =56.91 degrees. We then have to solve for initial y component of velocity, so we have 29.3sin(56.91)=24.55. We then say, since the graph has 5 seconds at the end point: s=.5(16+16)*5=80m. Is that right? for a. And then is b just s=.5(24.546+0)*2.5=30.6825? Thanks for the help, I just am unsure. Thanks and have great day :)
 
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  • #2
The image of the graph is attached below:
 

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  • #3
Looks right to me.
 

FAQ: X and Y components of a vector using a graph of speed/time

1. What is the difference between "X" and "Y" components of a vector?

The "X" and "Y" components of a vector refer to the horizontal and vertical parts of the vector, respectively. They represent the magnitude and direction of the vector in two dimensions.

2. How are the "X" and "Y" components of a vector calculated?

The "X" and "Y" components of a vector can be calculated using trigonometric functions. The "X" component can be found by multiplying the magnitude of the vector by the cosine of the angle between the vector and the horizontal axis. The "Y" component can be found by multiplying the magnitude of the vector by the sine of the angle.

3. What does the graph of speed/time for "X" and "Y" components of a vector represent?

The graph of speed/time for "X" and "Y" components of a vector shows the change in the magnitude of the vector over time in the horizontal and vertical directions. It can also show the direction of the vector at different points in time.

4. How do you interpret the slope of the graph for "X" and "Y" components of a vector?

The slope of the graph for "X" and "Y" components of a vector represents the rate of change of the vector's magnitude over time. A steeper slope indicates a greater change in magnitude, while a flatter slope indicates a slower change.

5. Can the "X" and "Y" components of a vector be negative?

Yes, the "X" and "Y" components of a vector can be negative. A negative component indicates that the vector is pointing in the opposite direction of the positive axis in that direction. For example, a negative "X" component means the vector is pointing to the left, and a negative "Y" component means the vector is pointing downwards.

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