A rock thrown horizontally from the roof edge of a 12.4 m-high building hits the ground below, a horizontal distance of 17.8 m from the building. What is the overall speed of the rock when it hits the ground?

Answers

Answer 1

The overall speed of the rock when it hits the ground is 24.4 m/s.

We can solve this problem using kinematic equations of motion. Since the rock is thrown horizontally, its initial vertical velocity is zero.

Let's use the following kinematic equation to find the final velocity of the rock (v):

v² = u² + 2as

where u is the initial velocity (in this case, u = 0), a is the acceleration due to gravity (-9.81 m/s²), and s is the vertical distance the rock falls (12.4 m). Solving for v, we get:

v = sqrt(2as) = sqrt(2 x (-9.81 m/s²) x 12.4 m) = 17.26 m/s

Now that we have found the final vertical velocity, we can use it to find the time it takes for the rock to fall to the ground.

The time (t) can be found using the following kinematic equation:

s = ut + (1/2)at²

where s is the horizontal distance the rock travels (17.8 m), u is the horizontal velocity of the rock (which is constant), and a is the horizontal acceleration (which is zero). Since the initial horizontal velocity is equal to the final horizontal velocity, we can use the following equation to find u:

v = u

u = v = 17.26 m/s

Now we can plug in the known values to find t:

17.8 m = 17.26 m/s x t

t = 1.03 s

Finally, we can use the horizontal distance and time to find the horizontal velocity (v_h) using the equation:

v_h = s/t = 17.8 m / 1.03 s = 17.28 m/s

Therefore, the overall speed of the rock when it hits the ground is the vector sum of the horizontal and vertical velocities:

v_overall = sqrt(v_h² + v²) = sqrt((17.28 m/s)² + (17.26 m/s)²) = 24.4 m/s

So the overall speed of the rock when it hits the ground is 24.4 m/s.

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Related Questions

a student is 2.50m away from a convex lens while her image is 1.80m from the lens, what is the focal length?

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To find the focal length of a convex lens, we can use the formula:
1/f = 1/di + 1/do

Where f is the focal length, di is the distance of the image from the lens, and do is the distance of the object from the lens.

We are given that the student is 2.50m away from the lens, so do = 2.50m. We are also given that the image is 1.80m from the lens, so di = 1.80m.

Plugging these values into the formula, we get:
1/f = 1/1.80 + 1/2.50

Simplifying this equation, we get:
1/f = 0.5556

Multiplying both sides by f, we get:
f = 1.80 / 0.5556

Solving for f, we get:
f ≈ 3.24 meters

Therefore, the focal length of the convex lens is approximately 3.24 meters.

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A convex lens is 1.80 meters from a student who is 2.50 meters distant, and its focal length is 1.04 meters.

To solve this problem, we can use the lens equation:

1/f = 1/do + 1/di

where f is the focal length of the lens, do is the object distance (distance of the object from the lens), and di is the image distance (distance of the image from the lens).

In this problem, the object distance is do = 2.50 m and the image distance is di = 1.80 m. We can plug these values into the lens equation and solve for the focal length:

1/f = 1/do + 1/di

1/f = 1/2.50 + 1/1.80

1/f = 0.4 + 0.56

1/f = 0.96

f = 1/0.96

f ≈ 1.04 meters

Therefore, the focal length of the convex lens is approximately 1.04 meters.

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maxwell's equations are a complete description of electric and magnetic fields. how many equations are there?

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Maxwell's equations are a complete description of electric and magnetic fields. There are four equations in Maxwell's equations. These four equations are:

1. Gauss's Law for Electric Fields: Describes the relationship between electric charges and the electric field produced by them.
2. Gauss's Law for Magnetic Fields: States that there are no magnetic monopoles, and the magnetic field lines are always closed loops.
3. Faraday's Law of Electromagnetic Induction: Describes the induced electromotive force (EMF) in a closed circuit produced by a changing magnetic field.
4. Ampere's Law with Maxwell's Addition: Relates the magnetic field around a closed loop to the electric current passing through the loop and the rate of change of the electric field.

These four equations collectively provide a comprehensive description of electric and magnetic fields and their interactions.

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T/F : Staleness and burnout are not associated with overtraining.

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False. Staleness and burnout are often associated with overtraining, which occurs when an individual exceeds their capacity to recover from intense physical training or activity.

Overtraining can lead to physical and psychological symptoms, including decreased performance, fatigue, irritability, and decreased motivation. It is important for individuals to listen to their bodies and take rest and recovery periods to prevent overtraining and associated symptoms.

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consider the picture above of mars's orbit around the sun. which spot shows where mars will be when we see it in retrograde motion on earth?

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When retrograde motion occurs and how it is related to Mars's orbit around the Sun:

Retrograde motion occurs when a planet appears to move backward in the sky from Earth's perspective. In the case of Mars, this happens when Earth overtakes Mars in their respective orbits around the Sun.

To understand when Mars will be in retrograde motion, consider these steps:

1. Picture both Mars and Earth orbiting the Sun, with Mars having a larger, slower orbit due to its greater distance from the Sun.
2. As Earth moves faster in its orbit, it eventually catches up to and passes Mars.
3. During this time, the relative positions of Earth, Mars, and the Sun create the illusion of Mars moving backward in the sky, as seen from Earth.

So, when trying to identify the spot where Mars will be in retrograde motion, look for the point in its orbit where Earth is passing Mars, creating the optical illusion of Mars moving backward in the sky.

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5. Explain the law of conservation of energy using a relevant example from every day life.​

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The law of conservation of energy states that energy is neither created nor destroyed but is transformed from one form to another.

What is law of conservation of energy?

The law of conservation of energy is the law that states that energy is neither created nor destroyed but is transformed from one form to another.

Examples of activities of everyday life that shows the conservation of energy include the following:

For loudspeaker, electrical energy is converted into sound energy.

For a microphone, sound energy is converted into electrical energy.

For a generator, mechanical energy is converted into electrical energy.

When fuels are burnt, chemical energy is converted into heat and light energy

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An example of the law of conservation of energy is a roller coaster.

What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

A roller coaster car gains kinetic energy as it moves down the track, but it also loses potential energy. At the bottom of the track, the car has the most kinetic energy and the least potential energy, while at the top of the track, it has the most potential energy and the least kinetic energy. However, the total amount of energy in the system remains constant.

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A mass of 25. 0 kg is acted upon by two forces: is 15. 0 n due east and is 10. 0 n and due north. The acceleration of the mass is

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the acceleration of the mass is 0.7212 m/s^2.

To find the acceleration of the mass, we need to first determine the net force acting on it. We can do this by using vector addition to add the two forces together.

Using the Pythagorean theorem, we can find the magnitude of the diagonal force:

sqrt[[tex](15N)^{2}[/tex] + [tex](10N)^{2}[/tex]] = sqrt[225 + 100] = sqrt(325) = 18.03 N

The direction of this force can be found using the inverse tangent function:

theta =[tex]tan^{-1}(10.0N/15.0N)[/tex] = 33.69 degrees north of east

We can now use vector addition to find the net force on the mass:

F_net = sqrt[[tex](15N)^{2}[/tex] + [tex](10N)^{2}[/tex]] = 18.03 N, at an angle of 33.69 degrees north of east

To find the acceleration of the mass, we can use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration:

F_net = ma

Solving for the acceleration, we get:

a = F_net / m = 18.03 N / 25.0 kg = 0.7212 m/s^2

Therefore, the acceleration of the mass is 0.7212 m/s^2.

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