ball roll off table with speed 2ft/s the table is 3.5ft high. determine the point at which ball hits the floor and find its speed at instant of impact. 

To solve this problem, we need to approach it in two parts: 
1. Determining where the ball hits the floor (the horizontal distance from the table).
2. Finding the ball's speed at the instant of impact (combining horizontal and vertical components).

### Part 1: Time of flight (vertical motion)

The ball is rolling off a table that is 3.5 feet high with no initial vertical velocity (because it rolls off horizontally). The only force acting in the vertical direction is gravity.

Using the kinematic equation for vertical motion:
\[
y = v_{0y}t + \frac{1}{2}gt^2
\]
where:
- \(y = 3.5 \, \text{ft}\) (height of the table),
- \(v_{0y} = 0 \, \text{ft/s}\) (initial vertical velocity),
- \(g = 32.2 \, \text{ft/s}^2\) (acceleration due to gravity),
- \(t\) is the time in seconds.

Thus, the equation becomes:
\[
3.5 = 0 + \frac{1}{2}(32.2)t^2
\]

Solving for \(t\):
\[
t^2 = \frac{2(3.5)}{32.2} = 0.2174
\]
\[
t = \sqrt{0.2174} \approx 0.466 \, \text{seconds}
\]

So, the ball is in the air for approximately 0.466 seconds.

### Part 2: Horizontal distance (range)

The ball has a horizontal velocity of 2 ft/s, and it travels for 0.466 seconds before hitting the ground. The horizontal distance \(x\) is given by:
\[
x = v_{x} t
\]
where:
- \(v_{x} = 2 \, \text{ft/s}\),
- \(t = 0.466 \, \text{seconds}\).

Thus:
\[
x = 2 \times 0.466 \approx 0.932 \, \text{feet}
\]

The ball hits the ground approximately **0.932 feet** away from the table.

### Part 3: Speed at impact

At the moment of impact, the ball has both a horizontal velocity (\(v_{x}\)) and a vertical velocity (\(v_{y}\)). The horizontal velocity remains constant at 2 ft/s, while the vertical velocity increases due to gravity.

To find the vertical velocity at impact, we use the kinematic equation:
\[
v_{y} = v_{0y} + gt
\]
Since \(v_{0y} = 0\), we have:
\[
v_{y} = (32.2)(0.466) \approx 15.01 \, \text{ft/s}
\]

Now, to find the total speed at impact (\(v\)), we combine the horizontal and vertical components using the Pythagorean theorem:
\[
v = \sqrt{v_{x}^2 + v_{y}^2}
\]
\[
v = \sqrt{(2)^2 + (15.01)^2} \approx \sqrt{4 + 225.3} \approx \sqrt{229.3} \approx 15.14 \, \text{ft/s}
\]

### Final answers:
1. **The ball hits the ground approximately 0.932 feet away from the table.**
2. **The ball's speed at the instant of impact is approximately 15.14 ft/s.**

Would you like further details on any part of this solution? Here are some related questions to explore:

1. How would the result change if the initial speed of the ball was 3 ft/s instead of 2 ft/s?
2. What would happen if the height of the table was doubled?
3. How can air resistance affect the ball's motion in this scenario?
4. How would you calculate the total distance traveled by the ball, considering both vertical and horizontal motion?
5. Can you derive the time of flight equation if the ball had an initial vertical velocity?

**Tip:** When dealing with projectile motion, always break the motion into horizontal and vertical components for easier analysis.

A ball rolls off a table with a speed of 2 ft/s. The table is 3.5 ft high. Determine the point at which the ball hits the floor and find its speed at the instant of impact.

Learn how to calculate where a ball rolling off a 3.5-foot high table with a speed of 2 ft/s hits the ground and its speed at the moment of impact. This detailed solution covers projectile motion, including horizontal and vertical kinematics, suitable for students in Grades 9-12.

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