Question

record 206438@nhps.net as the email to be included with my response
name *
kimora downing ph
what happens to the potential energy of the ball as you increase the height from which it is dropped?
the potential energy stays the same
the potential energy decreases
the potential energy increases
the ball has no potential energy at any height

Explanation:

Step1: Recall the formula for gravitational potential energy

The formula for gravitational potential energy is \(U = mgh\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (\(g\approx9.8\ m/s^{2}\) near the Earth's surface), and \(h\) is the height of the object above a reference level.

Step2: Analyze the relationship between height and potential energy

Since \(m\) (mass of the ball, assuming it is constant) and \(g\) (acceleration due to gravity, approximately constant near the Earth's surface) are constants in this context, the potential energy \(U\) is directly proportional to \(h\) (height). Mathematically, if we consider two heights \(h_1\) and \(h_2\) such that \(h_2>h_1\), then \(U_1=mgh_1\) and \(U_2 = mgh_2\). Since \(g> 0\) and \(m>0\), when \(h\) increases (i.e., \(h_2 - h_1>0\)), \(U_2-U_1=mg(h_2 - h_1)>0\).

Answer:

The potential energy increases.