An illustrative image depicting a physics problem. Show a scene with a person with Black descent, male, standing at ground level, just released a rock into the air. The trajectory of the rock is visually indicated, shown at a height of 15.0 m in mid-flight, moving upward. The illustration seems to freeze the moment when the rock is 15.0 m above the ground, traveling upward. This image does not contain any text but is insightful about the described action.

You throw a 20-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is travelling at 25.0 m/s upward. Use the work-energy theorem to find a) its speed just as it left the ground; b) its maximum height.

knowns:

F=20N
h=15m
V=25m/s

unknowns:
max h
V1

work-energy theorem: W=K2-K1
K=.5*mV^2

so, m(-g)h=.5mV2^2-.5mV1^2

this simplifies to 2(-g)h=V2^2-V1^2

therefore, V1=sqrt(V2^2+2gh)

now all you have to do is plug in numbers.

V1=sqrt(25^2+2*9.8*15)=30.32m/s

max height is where V=0

we can use the work-energy theorem again.

this time we get: 2(-g)h=0-30.32^2

so, max h=-(30.32^2)/2(-g)=46.90m

the final answer is:
max height = 46.90m
and initial velocity = 30.32m/s

Thank you

Ook

Well, let's do some physics with a sprinkle of humor, shall we?

a) To find the rock's speed just as it left the ground, we can use the work-energy theorem. This theorem states that the work done on an object is equal to the change in its kinetic energy.

Since the rock is thrown vertically, we know that the work done on it is equal to the change in its gravitational potential energy. The initial kinetic energy of the rock is zero, so the work done on it is equal to the change in its potential energy.

Now, we can calculate it. The work done on an object is given by the formula: work = force × distance. Since the rock weighs 20 N, and it is 15.0 m above the ground, the work done on it is 20 N × 15.0 m = 300 J.

Since the work done is equal to the change in potential energy, we can rewrite it as 300 J = m × g × h (where m is the mass of the rock, g is the acceleration due to gravity, and h is the height).

Now, we can solve for the mass of the rock: m = 300 J / (9.8 m/s² × 15.0 m) ≈ 2.04 kg.

With this information, we can calculate the speed just as it left the ground. The initial kinetic energy can be found using the formula: kinetic energy = 0.5 × mass × speed².

Since the initial kinetic energy is equal to the work done on the rock, we can rewrite it as 300 J = 0.5 × 2.04 kg × speed².

Solving for speed gives us: speed = sqrt(300 J / (0.5 × 2.04 kg)) ≈ sqrt(294.12 m²/s²) ≈ 17.14 m/s.

So, the speed just as the rock left the ground is approximately 17.14 m/s. The rock definitely started off with a good push, not too shabby!

b) Now let's calculate the maximum height the rock reaches. We know that at the maximum height, the rock's velocity will be zero. Based on the work-energy theorem, we can say that the work done on the rock is equal to the change in its kinetic energy.

Since the velocity is zero at the maximum height, the change in the rock's kinetic energy is equal to its initial kinetic energy. We can use the formula: work = 0.5 × mass × velocity².

Now, we can solve for the change in potential energy since the work done is equal to the change in potential energy: change in potential energy = 0.5 × 2.04 kg × (0 m/s)² = 0 J.

With the change in potential energy known, we can calculate the maximum height using the formula: change in potential energy = mass × g × change in height.

Substituting the values, we have: 0 J = 2.04 kg × 9.8 m/s² × change in height.

Solving for the change in height gives us: change in height = 0 J / (2.04 kg × 9.8 m/s²) = 0 m.

Since the change in height is zero, the maximum height the rock reaches is also zero. Well, seems like our rock didn't make it very high. It's okay, though. Maybe it just preferred being close to the ground.

To find the speed at which the rock left the ground, we can use the work-energy theorem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy.

a) Since the only force acting on the rock is gravity, the net work done on the rock is equal to the work done against gravity. The work done against gravity can be calculated using the formula W = mgh, where W is the work done, m is the mass, g is the acceleration due to gravity, and h is the height.

Given that the height is 15.0 m and the rock has a mass of m, we can write the equation as:

W = mgh

The work done against gravity is equal to the change in kinetic energy, which is given by:

W = ΔKE

Since the rock starts from rest at ground level, its initial kinetic energy is zero. Therefore, we can write:

mgh = 0.5mv^2

Simplifying the equation, we get:

gh = 0.5v^2

Now we can substitute the known values:

(9.8 m/s^2)(15.0 m) = 0.5v^2

147 m = 0.5v^2

Dividing both sides by 0.5, we get:

294 m/s^2 = v^2

Taking the square root of both sides, we find:

v ≈ 17.14 m/s

Therefore, the speed at which the rock left the ground is approximately 17.14 m/s.

b) To find the maximum height reached by the rock, we can use the conservation of mechanical energy. At its maximum height, the rock has no kinetic energy but has gravitational potential energy.

The total mechanical energy, E, is given by:

E = KE + PE

Since the rock starts from rest, its initial kinetic energy is zero. Therefore, the total mechanical energy is equal to the gravitational potential energy at the maximum height:

E = PE

The gravitational potential energy is given by:

PE = mgh

Substituting the known values:

PE = (m)(9.8 m/s^2)(15.0 m)

The gravitational potential energy is equal to the total mechanical energy, so:

PE = E

Therefore, the maximum height reached by the rock is 1470 J.

To solve this problem using the work-energy theorem, we need to consider the different forms of energy involved and how they change throughout the motion of the rock.

a) To find the speed of the rock just as it left the ground, we need to equate the initial kinetic energy, denoted as KE_i, with the work done on the rock, denoted as W. According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy.

The initial kinetic energy of the rock is zero when it is at rest on the ground, so we have:

KE_i = 0

The work done on the rock is given by the formula:

W = ∆KE = KE_f - KE_i

where ∆KE is the change in kinetic energy, KE_f is the final kinetic energy, and KE_i is the initial kinetic energy.

From the problem statement, we know that the rock is thrown vertically, so its velocity just as it left the ground is equal to the velocity when it was 15.0 m above the ground (25.0 m/s upward). Thus, the final kinetic energy can be calculated using the formula:

KE_f = (1/2) * m * v^2

where m is the mass of the rock, and v is the final velocity.

Given that the mass of the rock is not provided, we can assume it to be 1 kg for simplicity.

Substituting the given values, we have:

KE_f = (1/2) * 1 kg * (25.0 m/s)^2 = 1/2 * 1 kg * 625 m^2/s^2 = 312.5 J

Now, the work done on the rock can be calculated as:

W = ∆KE = KE_f - KE_i = 312.5 J - 0 J = 312.5 J

Since the work done on the rock is equal to the initial kinetic energy, we can write:

W = KE_i

Therefore, the speed of the rock just as it left the ground can be found by equating the work done to the initial kinetic energy:

KE_i = 312.5 J

b) To find the maximum height reached by the rock, we need to consider the conservation of energy. At the maximum height, all of the initial kinetic energy is transformed into potential energy (PE_max), given by the formula:

PE_max = m * g * h

where m is the mass of the rock, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the maximal height.

Considering the conservation of energy, we can equate the initial kinetic energy (KE_i) to the potential energy at the maximum height (PE_max):

KE_i = PE_max

Since we have already found that KE_i = 312.5 J, we can solve for the maximal height (h):

312.5 J = m * 9.8 m/s² * h

However, we don't know the mass of the rock. This means that we can only determine the maximum height in terms of the mass or by assuming a value for the mass.

So, to summarize, the explanations above help to understand the approach needed to solve the problem, but the actual values for the speed just as it left the ground and the maximum height reached by the rock cannot be determined without knowing the mass of the rock or making an assumption about its value.