A ball is being shot at a moving hoop the curved line labeled b(s) maps the path of the ball and the line labeled h(s) maps the path of the hoop. How long does it take for the ball to go down through the hoop to the nearest second

Question

On graph a straight dotted line from starting at 8 going across with a point (17.906,8) then the second is a downward u shape starting at 5 crossing 1st line at 8 center is (10,13) down to (17.906,8) then down to (23,0)

Answer 1

To determine how long it takes for the ball to travel through the hoop based on the provided graph description, we interpret the positions and movements of the ball and the hoop.

Based on your description:

1. The straight line (let's call it line \(b(s)\)) starts at the height of 8 and intersects the downward U-shaped curve (let's call it line \(h(s)\)) at the coordinate (17.906, 8).

2. The U-shaped curve has its vertex at the point (10, 13), indicating its maximum height is at \(y = 13\) when \(x = 10\).

3. The U-shaped curve goes down from (10, 13) to the points (17.906, 8) and (23, 0).

Just examining the coordinates provided, it seems the ball is moving along line \(b(s)\) and intersects the hoop (line \(h(s)\)) at point (17.906, 8).

Finding the time taken for the ball to pass through the hoop

The key considerations to address the time it takes can be done through understanding the vertical distance the ball travels through the hoop from entry to exit:

1. Entry Point: At (17.906, 8) - this is where the ball enters the hoop.
2. Exit Point: The hoop appears to go down from (17.906, 8) further down to (23, 0).

Since the ball descends vertically from (17.906, 8) to (23, 0), the vertical distance is calculated as:

• Initial height: \(y_{initial} = 8\)
• Final height at exit: \(y_{final} = 0\)

Vertical distance calculation:

\[ \Delta y = y_{initial} - y_{final} = 8 - 0 = 8 \]

Next, if we need the time, we would generally have to know the vertical speed (or vertical velocity) of the ball \(v\) or the equation governing its motion. If you state that it is in free fall, assuming no air resistance, using gravity \(g\) (approximately \(9.81 , \text{m/s}^2\)), the time \(t\) it would take is given by:

\[ y = \frac{1}{2} g t^2 \]

We need to rearrange the equation to solve for time \(t\):

\[ 8 = \frac{1}{2} \cdot 9.81 \cdot t^2 \]

\[ 8 = 4.905 \cdot t^2 \]

\[ t^2 = \frac{8}{4.905} \approx 1.63 \]

\[ t \approx \sqrt{1.63} \approx 1.28 , \text{seconds} \]

Conclusion

Rounding to the nearest second, it takes approximately 1 second for the ball to pass through the hoop, assuming it is in free fall under gravity. If additional forces or velocities are at play, please provide those details for a more precise answer!

Answer 2

Options
8
18
17.9
17

Answer 3

To evaluate the options you provided, let's clarify and re-analyze the situation based on the points and the nature of the motion.

Given the previous information:

• The ball reaches the hoop at the point (17.906, 8) and exits downward to (23, 0).
• The hoop's vertex is at (10, 13).

From the earlier analysis, let's focus on how far the ball must drop after it enters the hoop at (17.906, 8) until it exits at (23, 0).

Re-evaluation of the Ball's Path:

1. Initial height upon entering hoop: \(y_{initial} = 8\) (at \(x = 17.906\))
2. Final height upon exiting the hoop: \(y_{final} = 0\) (at \(x = 23\))

The vertical distance it falls is:

\[ \Delta y = y_{initial} - y_{final} = 8 - 0 = 8 \]

Horizontal Distance

The ball moves horizontally from \(x = 17.906\) to \(x = 23\).

\[ \Delta x = 23 - 17.906 = 5.094 \]

Time Calculation

Assuming that the ball's speed horizontally is constant, if the initial vertical drop occurs under gravity, we can analyze its vertical motion separately:

Let's assume it falls straight down and we reapply the equations based on a free-fall motion:

Using the kinematic equation:

\[ \Delta y = v_{0y}t + \frac{1}{2}gt^2 \]

If we assume the ball starts dropping immediately (as it would from height 8), neglecting the forces acting on it other than gravity:

1. Set \(v_{0y} = 0\) (since we can assume it's at rest momentarily before falling vertically)

The equation simplifies to:

\[ 8 = 0 \cdot t + \frac{1}{2} \cdot 9.81 \cdot t^2 \] \[ 8 = 4.905 t^2 \] \[ t^2 = \frac{8}{4.905} \approx 1.63 \] \[ t \approx \sqrt{1.63} \approx 1.28 \text{ seconds} \]

Conclusion

The time to pass through the hoop would amount to approximately 1 second when considering only vertical motion.

However, based on the options you've given (8, 18, 17.9, 17), it seems these values represent timing at different segments or positions.

Given the options and the context of the problem, the nearest option would still logically connect back to the horizontal distance and sufficient time it's been given.

Reassessing based on all outlined context:

• If you're considering total time across distance or under some different metric, a reasonable interpretation would match to an approximation to the third option (17.9) generally leading to larger cumulative timeframes.

To conclude, among the options provided, I would select 17.9 seconds, as it is contextually closest to potential longer timeframes passed from a vertical/horizontal crossing motion perspective.