Question

this diagram shows a baseballs motion. which statement is best supported by the diagram? kinetic energy becomes potential energy between points r and s. maximum potential energy is at point p. maximum potential energy is at point r. potential energy becomes kinetic energy between points p and q.

Explanation:

Step1: Analyze the relationship between kinetic and potential energy

The total mechanical energy \(E = K + U\) (where \(K\) is kinetic energy and \(U\) is potential energy). Kinetic energy \(K=\frac{1}{2}mv^{2}\) (depends on speed \(v\)) and potential energy \(U = mgh\) (depends on height \(h\), \(m\) is mass, \(g\) is acceleration due to gravity).

Step2: Check each option

• Option 1: Between \(R\) and \(S\), the ball is moving down - height decreases. As \(h\) decreases, \(U=mgh\) decreases and \(K=\frac{1}{2}mv^{2}\) increases (because speed increases as it falls). So, potential energy becomes kinetic energy, not the other way around.
• Option 2: At point \(P\), the ball is at a lower height compared to point \(R\). Using \(U = mgh\), since \(h_{P}
• Option 3: At point \(R\), the ball is at the highest position in the given motion. Using \(U = mgh\) (where \(m\) is mass of the ball, \(g\) is acceleration due to gravity and \(h\) is height). Since \(h\) is maximum at \(R\) (compared to \(P\), \(Q\), \(S\), \(T\)), \(U = mgh\) is maximum at \(R\).
• Option 4: Between \(P\) and \(Q\), the ball is moving up - height increases. As \(h\) increases, \(U=mgh\) increases and \(K = \frac{1}{2}mv^{2}\) decreases (because speed decreases as it moves up). So, kinetic energy becomes potential energy, not the other way around.

Answer:

Maximum potential energy is at point \(R\).