At the local playground a child on a swing has a speed of 2.14 m/s when the swing is at its lowest point.

(a) To what maximum vertical height does the child rise, assuming he sits still and "coasts"? Ignore air resistance.

(b) To what maximum vertical height does the child rise if the speed is reduced to 1/5 the initial speed?

(a) The maximum vertical height the child rises is equal to the kinetic energy of the swing at its lowest point, divided by the gravitational force. This is equal to (2.14 m/s)^2 / (2 * 9.8 m/s^2) = 0.44 m.

(b) The maximum vertical height the child rises is equal to the kinetic energy of the swing at its lowest point, divided by the gravitational force. This is equal to (1/5 * 2.14 m/s)^2 / (2 * 9.8 m/s^2) = 0.044 m.

To find the maximum vertical height the child rises, we can use the principle of conservation of mechanical energy.

(a) When the swing is at its lowest point, it has its maximum kinetic energy. At the highest point, the swing has its maximum potential energy. The sum of kinetic energy and potential energy is constant if we ignore air resistance.

The formula for kinetic energy (KE) is given by KE = (1/2) * mass * velocity^2
The formula for potential energy (PE) is given by PE = mass * gravity * height

Let's assume the mass of the child is m, and the gravity is g.

At the lowest point, the swing has its maximum kinetic energy:
KE1 = (1/2) * m * (2.14 m/s)^2

At the highest point, the swing has its maximum potential energy:
PE2 = m * g * height

Since the total mechanical energy is constant:
KE1 + PE2 = KE2 + PE1

Since the child sits still and "coasts" without using any external force, the final kinetic energy (KE2) is zero:
KE1 + PE2 = 0 + PE1

Therefore, we can solve for the maximum vertical height (height) using the equation:
m * g * height = (1/2) * m * (2.14 m/s)^2

Simplifying:
g * height = (1/2) * (2.14 m/s)^2

Now we can plug in the values for g and solve for the height:
g = 9.8 m/s^2 (acceleration due to gravity)
height = [(1/2) * (2.14 m/s)^2] / 9.8 m/s^2

Calculating:
height ≈ 0.2312 m

Therefore, the child maximum vertical height the child rises when sitting still and coasting is approximately 0.2312 meters.

(b) Now let's consider the case where the speed is reduced to 1/5 of the initial speed. Let's say the new speed is v.

Using the same principle of conservation of mechanical energy, we can set up the equation:

(1/2) * m * (1/5 * v)^2 + m * g * height = (1/2) * m * (2.14 m/s)^2

Now, we can solve for the maximum vertical height (height):

(1/2) * m * (1/5 * v)^2 + m * g * height = (1/2) * m * (2.14 m/s)^2

Simplifying:
(1/2) * m * (1/25) * v^2 + m * g * height = (1/2) * m * (2.14 m/s)^2

Canceling out the mass:
(1/2) * (1/25) * v^2 + g * height = (1/2) * (2.14 m/s)^2

Simplifying further:
(1/50) * v^2 + g * height = (1/2) * (2.14 m/s)^2

We can solve for height by substituting the values of g and v:

g = 9.8 m/s^2 (acceleration due to gravity)
v = 2.14 m/s * (1/5) = 0.428 m/s

Substituting the values:
(1/50) * (0.428 m/s)^2 + 9.8 m/s^2 * height = (1/2) * (2.14 m/s)^2

Calculating:
(1/50) * (0.428 m/s)^2 + 9.8 m/s^2 * height ≈ 0.2294 m/s^2

Simplifying further:
0.000857 m^2/s^2 + 9.8 m/s^2 * height ≈ 0.2294 m/s^2

Now we can solve for height:
9.8 m/s^2 * height ≈ 0.2294 m/s^2 - 0.000857 m^2/s^2
height ≈ (0.2294 m/s^2 - 0.000857 m^2/s^2) / 9.8 m/s^2

Calculating:
height ≈ 0.0233 m

Therefore, the child's maximum vertical height when the speed is reduced to 1/5 of the initial speed is approximately 0.0233 meters.

To find the maximum vertical height the child rises in both scenarios, we can use the principle of conservation of mechanical energy.

The total mechanical energy of a system is the sum of its kinetic energy and potential energy at any given point. In the absence of air resistance, the mechanical energy remains constant throughout the swing.

Let's start with part (a):

(a) To find the maximum vertical height when the child is coasting at a speed of 2.14 m/s, we first need to determine the initial and final points in the swing path.

At the lowest point of the swing, the child has maximum kinetic energy and minimum potential energy. At the highest point, the child has maximum potential energy and minimum kinetic energy.

According to the principle of conservation of mechanical energy, the initial mechanical energy is equal to the final mechanical energy.

At the lowest point:
- Potential energy (PE) = 0, as the child is at the lowest height.
- Kinetic energy (KE) = 1/2 * mass * (speed)^2.

At the highest point:
- Potential energy (PE) is maximum.
- Kinetic energy (KE) = 0, as the child is momentarily at rest.

Since the mass of the child is not given in the question, it does not influence the relative heights reached during the swing.

Using the principle of conservation of mechanical energy, we equate the initial and final mechanical energy:

KE at lowest point = PE at highest point

1/2 * m * (speed at lowest point)^2 = m * g * h

Here, m is the mass of the child, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the maximum vertical height reached by the child.

Hence, to solve for the maximum height, we rearrange the equation and solve for h:

h = (speed at lowest point)^2 / (2 * g)

Now, let's calculate the maximum vertical height the child rises:

h = (2.14 m/s)^2 / (2 * 9.8 m/s^2)

Simplifying the expression:

h = 4.5796 m^2/s^2 / 19.6 m/s^2

h ≈ 0.2339 m

Therefore, the child rises to a maximum vertical height of approximately 0.2339 meters when coasting at a speed of 2.14 m/s.

Moving on to part (b):

(b) To find the maximum vertical height when the speed is reduced to 1/5 the initial speed, we need to apply the same approach as in part (a) but use the new speed value.

Let's denote the new speed as v_new.

Following the principle of conservation of mechanical energy, we can set up the equation:

1/2 * m * (v_new)^2 = m * g * h

Simplifying the equation:

h = (v_new)^2 / (2 * g)

Now, we can substitute the values and calculate the maximum vertical height the child rises:

h = (1/5 * 2.14 m/s)^2 / (2 * 9.8 m/s^2)

h = (0.428 m/s)^2 / 19.6 m/s^2

h ≈ 0.0092 m

Therefore, when the child's speed is reduced to 1/5 the initial speed (approximately 0.428 m/s), the child rises to a maximum vertical height of approximately 0.0092 meters.