Question

a pendulum is raised to a certain height and released from point a, as shown in the image below. at its release, the pendulum is also given an initial velocity of 14 m/s. assuming that the effects of friction and air resistance can be ignored, what will be the maximum height that the pendulum can reach - that is, what is the height at point b? (recall that ( g = 9.8 , \text{m/s}^2 ))

Explanation:

Step1: Apply conservation of mechanical energy

The total mechanical energy at point \(A\) (\(E_A\)) is the sum of kinetic energy (\(K_E=\frac{1}{2}mv^{2}\)) and potential energy (\(P_E = mgh\)). At point \(B\), the pendulum has only potential energy (\(E_B=mgh_B\)) since at maximum - height its velocity \(v = 0\). By the law of conservation of energy \(E_A=E_B\), so \(\frac{1}{2}mv_{A}^{2}+mgh_{A}=mgh_{B}\).

Step2: Simplify the energy - conservation equation

Divide the entire equation \(\frac{1}{2}mv_{A}^{2}+mgh_{A}=mgh_{B}\) by \(m\) (mass \(m\) cancels out because \(m
eq0\)). We get \(\frac{1}{2}v_{A}^{2}+gh_{A}=gh_{B}\).

Step3: Solve for \(h_B\)

Rearrange the equation \(\frac{1}{2}v_{A}^{2}+gh_{A}=gh_{B}\) for \(h_B\). First, express \(h_B=\frac{v_{A}^{2}}{2g}+h_{A}\). Given \(v_{A} = 14\ m/s\), \(g = 9.8\ m/s^{2}\), and \(h_{A}=20\ m\). Substitute the values: \(\frac{14^{2}}{2\times9.8}+20\). Calculate \(\frac{196}{19.6}+20\). Since \(\frac{196}{19.6}=10\), then \(h_B=10 + 20\).

Answer:

\(h_B = 30\ m\), so the answer is D.