Let's break down each part of the problem step by step:

Part a: How long is the ball in flight before it hits the floor? (7 points)

We will use the kinematic equation for vertical motion, where the ball is in free fall under the influence of gravity:

$y = v_{i_y} t + \frac{1}{2} g t^2$

• $y$ is the vertical displacement (1.4 m, since the table is 1.4 meters high),
• $v_{i_y} = 0 \, \text{m/s}$ (because the ball is initially moving only horizontally, so there is no initial vertical velocity),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity).

The equation simplifies to:

$1.4 = \frac{1}{2} \times 9.8 \times t^2$

Solving for $t$:

$t^2 = \frac{2 \times 1.4}{9.8} = \frac{2.8}{9.8} \approx 0.286$ $t \approx \sqrt{0.286} \approx 0.535 \, \text{seconds}$

So, the ball is in flight for approximately 0.54 seconds.

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Part b: How far has the ball gone horizontally from the table edge, when it hits the floor? (5 points)

The horizontal velocity remains constant throughout the flight, as there is no horizontal acceleration (neglecting air resistance). The distance $x$ the ball travels horizontally is given by:

$x = v_x t$

Where:

• $v_x = 2.6 \, \text{m/s}$ (the horizontal velocity),
• $t = 0.54 \, \text{seconds}$ (the time in flight, calculated above).

$x = 2.6 \times 0.54 \approx 1.404 \, \text{meters}$

So, the ball travels approximately 1.40 meters horizontally before hitting the floor.

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Part c: What is the ball's speed just before impact with the floor? (7 points)

To find the speed just before impact, we need to find both the horizontal and vertical components of velocity at that moment.

1. The horizontal velocity $v_x$ remains constant at $2.6 \, \text{m/s}$.
2. The vertical velocity $v_y$ just before impact can be calculated using the kinematic equation:

$v_y = v_{i_y} + g t = 0 + 9.8 \times 0.54 \approx 5.29 \, \text{m/s}$

Now, we can find the total speed just before impact using the Pythagorean theorem:

$v = \sqrt{v_x^2 + v_y^2}$ $v = \sqrt{(2.6)^2 + (5.29)^2} = \sqrt{6.76 + 27.98} = \sqrt{34.74}$ $v \approx 5.89 \, \text{m/s}$

So, the ball's speed just before impact is approximately 5.89 m/s.

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Part d: Sketch graphs of $x$ vs. $t$ and $v_y$ vs. $t$ (6 points)

• Graph of $x$ vs. $t$: This will be a straight line with a positive slope, because the horizontal velocity is constant, and $x = v_x t$. The slope of this line will be $2.6 \, \text{m/s}$, and the ball will have traveled 1.4 meters by the time $t = 0.54 \, \text{seconds}$.

• Graph of $v_y$ vs. $t$: This will be a straight line with a positive slope starting at $v_y = 0 \, \text{m/s}$ and increasing linearly with time, since the vertical velocity increases uniformly due to gravity. The slope of this line will be $9.8 \, \text{m/s}^2$, and at $t = 0.54 \, \text{seconds}$, $v_y$ will have reached $5.29 \, \text{m/s}$.

Summary of Answers:

1. Time in flight: 0.54 seconds.
2. Horizontal distance: 1.40 meters.
3. Speed just before impact: 5.89 m/s.
4. Graphs of $x$ vs. $t$ and $v_y$ vs. $t$ (as described).

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Follow-up questions:

1. What would happen to the flight time if the table height doubled?
2. How would the horizontal distance change if the horizontal speed were halved?
3. What effect does air resistance have on the ball’s motion?
4. How would the graphs change if the ball were dropped instead of rolling off the table?
5. What is the maximum height the ball reaches if it is initially thrown at an upward angle?

Tip: The horizontal and vertical motions are independent of each other in projectile motion.