a parent pushes a kid on a swing,

A) If the child on the swing rises 2m above the lowest point during each cycle, what is her speed at the lowest point of the swing?
B)If the beginning of a cycle (2m above the lowest point) the child is given a push so that her starting speed is 3 meters/second, what will be her speed at the bottom of the swing?
How do I approach the first question without the length of the rope or the weight of the child?

To approach the first question without knowing the length of the rope or the weight of the child, we can use the conservation of energy principle. The total mechanical energy of the system (child and swing) remains constant throughout the motion.

At the highest point of the swing (2m above the lowest point), the kinetic energy is zero since the child is momentarily at rest. All the mechanical energy is in the form of potential energy.

At the lowest point of the swing, all the potential energy is converted to kinetic energy. Since the child is at her lowest point, the potential energy is zero. Therefore, all of the mechanical energy is in the form of kinetic energy.

Since the potential energy is proportional to the square of the speed and the kinetic energy is given by (1/2)mv^2, we can set up the following equation:

(1/2)mv^2 = mgh

Where m is the mass of the child (which cancels out both sides of the equation), v is the speed at the lowest point, and h is the height above the lowest point (equal to 2m in this case).

Simplifying the equation, we have:

(1/2)v^2 = 2gh

Since we don't have values for m or h, we don't need them to find the speed at the lowest point. We can assume that m and h are constant and cancel them out to get:

(1/2)v^2 = 2g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Solving for v, we get:

v^2 = 4g

v = √(4g)

v ≈ 2√g

Therefore, the speed at the lowest point of the swing is approximately 2 times the square root of the acceleration due to gravity.