Question

A stone falls from a cliff and travels $24.5m$ in the last second before it reaches the ground at the foot of the clif. Find the height of the cliff. ($g=9.8m{s}^{-2}$)

Answer 1

Step 1: Given

Distance traveled in the last second = $24.5m$

Acceleration due to gravity $g=9.8m{s}^{-2}$

Step 2: Formula

$$\begin{gathered} s_n=u+g(n-\frac{1}{2}) \\ s=ut+\frac{1}{2}g{t}^2 \end{gathered}$$

Step 3: Calculation of time

$$\begin{gathered} s_n=u+g(n-\frac{1}{2}) \\ 24.5=0+9.8(n-\frac{1}{2}) \\ 24.5=9.8n-4.9 \\ 9.8n=24.5+4.9 \\ n=\frac{29.4}{9.8} \\ n=3 \end{gathered}$$

So time taken to reach the ground is $3s$.

Step 4: Calculation of height

$$\begin{gathered} s=ut+\frac{1}{2}g{t}^2 \\ =0\times 3+\frac{1}{2}\times 9.8\times 3^2 \\ =4.9\times 9 \\ =44.1m \end{gathered}$$

So, the height of the cliff is $44.1m$.