A swing sits 5 m above the ground when it is still. John lifts the swing and releases it. At its lowest point, its velocity is measured to be 5.53 m/s. How high above the ground did John lift the swing?

To find out how high above the ground John lifted the swing, we can use the principle of conservation of energy.

At its highest point (when it is still), the swing has only potential energy. As it swings down to its lowest point, all of its potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at the highest point to the kinetic energy at the lowest point.

The potential energy (PE) of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the ground.

The kinetic energy (KE) of an object is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.

Since the mass of the swing is not given, we can assume it cancels out in the equation. Therefore, we can write:

PE = KE

mgh = (1/2)mv^2

gh = (1/2)v^2

Substituting the given values:

9.8 m/s^2 * h = (1/2) * (5.53 m/s)^2

9.8 m/s^2 * h = (1/2) * (5.53 m/s)^2

9.8 m/s^2 * h = 15.244 m^2/s^2

h = 15.244 m^2/s^2 / (9.8 m/s^2)

h ≈ 1.557 m

Therefore, John lifted the swing to a height of approximately 1.557 meters above the ground.