This is what I have to do. I dont get it, I've tried and tried. Contacted my teacher and still dont get it. Please help. I'm frustrated.

The projectile’s mass is not needed.
 The projectile must be launched on the horizontal plane.
 The earth drops 8 inches for every mile traveled.
 The acceleration due to gravity is 9.81 meters per second squared.
 Ignore air resistance.
You will have to calculate how far an object drops every second due to gravity

How fast does an object need to travel to stay in orbit for two seconds?
How fast does an object need to travel to stay in orbit for three seconds?

Please be careful to state the problem for us the way it was presented to you.

I am sure your physics teacher did not say that this projectile was "in orbit" because it is not. I also do not trust the mixture of inches and meters.
Anyway the horizontal velocity is constant, period. Call it u
Now how far does it fall in two seconds?
h = (1/2)g t^2 = (1/2)(9.81)(4) = call it h in meters (about 20 meters but you calculate)
now how far is that away at the given slope of (8/5280*12)
d = 20 (5280*12/8)
so the speed is d meters/ 2 seconds

Thanks

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To calculate the speed required for an object to stay in orbit for a specific amount of time, we need to consider a few principles:

1. Centripetal Force: In order to stay in orbit, an object must experience a centripetal force that balances the force of gravity pulling it towards the Earth's center.

2. Centripetal Force Formula: The centripetal force can be calculated using the formula: F_c = (m * v^2) / r, where F_c is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the orbit.

3. Equating Forces: The centripetal force must be equal to the gravitational force, which is given by F_g = (m * g), where g is the acceleration due to gravity.

4. Simplification: Since we are asked to calculate the speed required for an object to stay in orbit for a certain amount of time, we can simplify the problem by assuming a circular orbit and equating the time it takes to complete one orbit with the desired time (2 or 3 seconds).

Now, let's move on to calculating the velocities:

1. Two-Second Orbit:
- The time it takes to complete one orbit is 2 seconds.
- To calculate the radius of the orbit, we will use the information given: "The earth drops 8 inches for every mile traveled." Since 1 mile is equal to 5280 feet and 1 foot is equal to 12 inches, we have 8/5280/12 ≈ 5.36 x 10^(-5) feet of drop per foot of travel.
- In 2 seconds, an object in orbit travels a distance equal to the circumference of the orbit. The circumference can be calculated using the formula: C = 2 * π * r, where π is approximately equal to 3.14159.
- Equating the distance traveled with the drop per foot of travel, we have: 2 * π * r ≈ 5.36 x 10^(-5) * r.
- Solving for the radius, we get: r ≈ (5.36 x 10^(-5)) / (2 * π) ≈ 8.53 x 10^(-6) feet.
- Now, we can calculate the velocity using the centripetal force formula: (m * v^2) / r = m * g.
- Since the mass of the object is not given, we can cancel it out from both sides of the equation.
- The equation becomes: v^2 / r = g.
- Solving for the velocity, we have: v^2 ≈ g * r.
- Plugging in the values, we get: v^2 ≈ (9.81 m/s^2) * (8.53 x 10^(-6) feet).
- Converting the acceleration due to gravity to feet per second squared, we have: v^2 ≈ (9.81 * 3.28084) * 8.53 x 10^(-6) feet.
- Simplifying further, we get: v^2 ≈ 0.0322977 feet^2/s^2.
- Finally, taking the square root of both sides, we find: v ≈ 0.17991 feet/s.

Therefore, an object needs to travel at a speed of approximately 0.17991 feet per second to stay in orbit for two seconds.

2. Three-Second Orbit:
- Following the same steps as above, we find that an object needs to travel at a speed of approximately 0.25856 feet per second to stay in orbit for three seconds.

I hope this explanation helps you understand the process of calculating the required speeds for objects to stay in orbit for specific times. If you have any further questions or need additional clarification, please feel free to ask.