Question

read the scenario, then answer the questions. a 2 kg ball is thrown upward with a velocity of 15 m/s. what is the kinetic energy of the ball as it is being thrown? what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?

Explanation:

Step1: Calculate kinetic energy

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\), where \(m = 2\space kg\) and \(v=15\space m/s\).
Substitute the values: \(K=\frac{1}{2}\times2\times15^{2}\)

Step2: Simplify the expression

\(K = 1\times225=225\space J\)

Step3: Use conservation of mechanical energy

At the maximum - height, the velocity \(v = 0\space m/s\), so the kinetic energy \(K = 0\space J\). According to the conservation of mechanical energy \(E = K + U\) (where \(E\) is the total mechanical energy, \(K\) is kinetic energy, and \(U\) is potential energy). Initially, \(E=K_{initial}+U_{initial}\), and at the maximum height \(E = K_{max - height}+U_{max - height}\). Assuming \(U_{initial}=0\) (taking the initial height as the reference level), then \(U_{max - height}=K_{initial}\)

Answer:

The kinetic energy as it is being thrown is \(225\space J\). The potential energy at the maximum height is \(225\space J\)