Question

A stone is thrown downward from the top of a cliff at 24m/s and hits the ground 7 seconds later. How tall is the cliff?​

Answer 1

The height of the cliff is 408.1 m

Explanation:

To find the height of the cliff given a stone is thrown downward from the top of a cliff at 24m/s and hits the ground 7 seconds later, use constant acceleration equations (SUVAT).

Constant Acceleration Equations (SUVAT)

`\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+(1)/(2)at^2\\\\ s&=\left((u+v)/(2)\right)t\\\\v^2&=u^2+2as\\\\s&=vt-(1)/(2)at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^(-1)$\\\\$v$ = final velocity in ms$^(-1)$\\\\$a$ = acceleration in ms$^(-2)$\\\\$t$ = time in s (seconds)\end{minipage}}`

Note: When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

List the given variables, taking downwards as positive:

• s = s
• u = 24 m/s
• a = 9.8 m/s²
• t = 7 s

Select the SUVAT equation with s, u, a and t in it:

`s=ut+(1)/(2)at^2`

`\begin{aligned}\textsf{Substitute the values}: \quad s&=24(7)+(1)/(2)(9.8)(7)^2\\s&=168+240.1\\s&=408.1\; \sf m \end{aligned}`

Therefore, the height of the cliff is 408.1 m.