Two spheres are launched horizontally from a 1.0 -high table. Sphere is launched with an initial speed of 4.5 . Sphere is launched with an initial speed of 3.0 .

What is the time for the sphere A to hit the floor?
What is the time for the sphere B to hit the floor?
What is the distance that sphere A travels from the edge of the table?
What is the distance that sphere B travels from the edge of the table?

Well, well, well, we have some falling spheres today! Let's see what we can do.

For sphere A, if we neglect air resistance (because it's not as fun as resistance, right?), we can use the good old kinematic equations.

The formula we need here is:
h = v0t + (1/2)gt^2, where
h is the height (1.0 meters), v0 is the initial speed (4.5 m/s), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Now, we need to find the time it takes for sphere A to hit the floor, so we'll just set h to 0:

0 = (4.5)t - (1/2)9.8t^2

Solving this quadratic equation (let me just dust off my math skills), we find that t is approximately 0.92 seconds for sphere A to hit the floor.

Now, for sphere B, we can go through the same process:

0 = (3.0)t - (1/2)9.8t^2

Solving for t again, we find that t is approximately 0.61 seconds for sphere B to make its graceful descent.

As for the distances traveled, we need to remember that these spheres were launched horizontally. So the horizontal distance traveled by a projectile can be found using the formula:

d = v0x * t, where v0x is the initial horizontal velocity and t is the time.

For sphere A, since it was launched horizontally, v0x is 4.5 m/s. So, using the previously determined time of 0.92 seconds:

d = 4.5 * 0.92 = 4.14 meters.

For sphere B, with an initial horizontal velocity of 3.0 m/s:

d = 3.0 * 0.61 = 1.83 meters.

So, to sum it up:
- Sphere A takes around 0.92 seconds to hit the floor.
- Sphere B takes around 0.61 seconds to hit the floor.
- Sphere A travels approximately 4.14 meters from the edge of the table.
- Sphere B travels about 1.83 meters from the edge of the table.

Hope I could amuse you while delivering the answers! Calculus can sometimes be a real joke!

To find the time it takes for each sphere to hit the floor, we can use the kinematic equation for vertical motion:

\[d = v_0t + \frac{1}{2}gt^2\]

Where:
- \(d\) is the vertical distance traveled (in this case, the height of the table, which is 1.0 m)
- \(v_0\) is the initial vertical velocity (which is 0 for both spheres, since they are launched horizontally)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s\(^2\))
- \(t\) is the time it takes for the sphere to hit the floor (this is what we're trying to find)

For Sphere A:
\(v_0 = 4.5\) m/s
Using the equation, we can rearrange it to solve for \(t\):

\[1.0 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2\]

Simplifying this equation, we get:

\[4.9t^2 = 1.0\]
\[t^2 = \frac{1.0}{4.9}\]
\[t = \sqrt{\frac{1.0}{4.9}}\]

Using a calculator, we can find that \(t \approx 0.452\) seconds.

For Sphere B:
\(v_0 = 3.0\) m/s
Following the same steps as above, we can find the time it takes for Sphere B to hit the floor:

\[1.0 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2\]
\[4.9t^2 = 1.0\]
\[t^2 = \frac{1.0}{4.9}\]
\[t = \sqrt{\frac{1.0}{4.9}}\]

Also, for Sphere B, \(t \approx 0.452\) seconds.

Therefore, both Sphere A and Sphere B take approximately 0.452 seconds to hit the floor.

To find the distance traveled by each sphere from the edge of the table, we can use the equation for horizontal motion:

\[d = v \cdot t\]

Where:
- \(d\) is the horizontal distance traveled
- \(v\) is the initial horizontal velocity
- \(t\) is the time it takes for the sphere to hit the floor (as calculated above)

For Sphere A:
\(v = 4.5\) m/s (since it is launched horizontally with an initial speed of 4.5 m/s)
\(t \approx 0.452\) seconds (as calculated above)

Using the equation, we can calculate the distance traveled:

\[d = 4.5 \cdot 0.452\]
\[d \approx 2.034\) meters

Therefore, Sphere A travels approximately 2.034 meters from the edge of the table.

For Sphere B:
\(v = 3.0\) m/s (since it is launched horizontally with an initial speed of 3.0 m/s)
\(t \approx 0.452\) seconds (as calculated above)

Using the equation, we can calculate the distance traveled:

\[d = 3.0 \cdot 0.452\]
\[d \approx 1.356\) meters

Therefore, Sphere B travels approximately 1.356 meters from the edge of the table.