Question

1. an object is dropped from a cliff of unknown height. assume the process of energy transformation is only 67% efficient:

a. what is the height of the cliff if the object is falling at a speed of 18 m/s just before hitting the ground? 3 marks - a
b. how high would the cliff be instead, if the object is falling at 18 m/s, but the efficiency is 100%? 3 marks - a

Explanation:

Step1: Apply energy conservation for part a

The potential energy at the top of the cliff \(U = mgh\) is converted to kinetic energy \(K=\frac{1}{2}mv^{2}\). Given efficiency \(\eta = 0.67\), so \(\eta mgh=\frac{1}{2}mv^{2}\). Cancel out \(m\) (mass of the object), then \(h=\frac{v^{2}}{2g\eta}\). Substitute \(v = 18m/s\), \(g = 9.8m/s^{2}\), \(\eta=0.67\)

$$h=\frac{18^{2}}{2\times9.8\times0.67}$$

$$h=\frac{324}{13.132}$$

Step2: Calculate the value for part a

$$h\approx24.7m$$

Step3: Apply energy conservation for part b

When \(\eta = 1\) (100% efficiency), from \(mgh=\frac{1}{2}mv^{2}\). Cancel out \(m\), then \(h=\frac{v^{2}}{2g}\). Substitute \(v = 18m/s\), \(g = 9.8m/s^{2}\)

$$h=\frac{18^{2}}{2\times9.8}$$

$$h=\frac{324}{19.6}$$

Step4: Calculate the value for part b

$$h\approx16.5m$$

Answer:

a. The height of the cliff is approximately \(24.7m\)
b. The height of the cliff is approximately \(16.5m\)