Question

1. a stone is dropped from a cliff and hits the ground 4.5 seconds later. how high is the cliff?

Explanation:

Step1: Identify the relevant formula

The formula for the height \(h\) of an object in free - fall is \(h = v_0t+\frac{1}{2}gt^2\), where \(v_0\) is the initial velocity, \(t\) is the time, and \(g\) is the acceleration due to gravity. Since the stone is dropped (\(v_0 = 0\)) and \(g = 9.8\ m/s^2\), the formula simplifies to \(h=\frac{1}{2}gt^2\).

Step2: Substitute the values

We know that \(t = 4.5\ s\) and \(g = 9.8\ m/s^2\). Substitute these values into the formula \(h=\frac{1}{2}\times9.8\times(4.5)^2\).
First, calculate \((4.5)^2=20.25\). Then, \(\frac{1}{2}\times9.8 = 4.9\).
So, \(h=4.9\times20.25\).

Step3: Calculate the result

\(h = 4.9\times20.25=99.225\ m\)

Answer:

\(99.225\ m\)