Question

A stone falls from rest from the top of a cliff,a second stone is thrown do wnward from the same height 2sec later with an initial speed of 30m/s.If both stones hit the ground below simultaneously,how high is the cliff?

Answer 1

Answer:

The cliff is 12.632 meter high.

Explanation:

Each stone experiments a free fall motion, that is, an uniformly accelerated motion due to gravity. We construct the respective equations of motion for each stone:

First stone

$y_{1} = y_{o,1} +v_{o,1}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Second stone

$y_{2} = y_{o,2} +v_{o,2}\cdot (t-2) +\frac{1}{2}\cdot g\cdot (t-2)^{2}$ (2)

Where:

$y_{1}$, $y_{2}$ - Final height of the first and second stone, in meters.

$y_{o,1}$, $y_{o,2}$ - Initial height of the first and second stone, in meters.

$v_{o,1}$, $v_{o,2}$ - Initial speed of the first and second stone, in meters per second.

$t$ - Time, in seconds.

$g$ - Gravitational acceleration, in meters per square second.

If we know that $y_{o,1} = y_{o,2}$, $y_{1} = y_{2} = 0\,m$, $v_{o,1} = 0\,\frac{m}{s}$, $v_{o,2} = -30\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$, then we find that time when both stones hit the ground simultaneously is:

$4.904\cdot t^{2} = -30\cdot (t-2)+4.904\cdot (t-2)^{2}$

$4.904\cdot t^{2} = -30\cdot t + 60 +4.904\cdot (t^{2}-4\cdot t +4)$

$-30\cdot t +60 -19.616\cdot t +19.616 = 0$

$49.616\cdot t = 79.616$

$t = 1.605\,s$

The height of the cliff is:

$y_{1} = y_{o,1} +v_{o,1}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}$

$y_{o,1} = 12.632\,m$

The cliff is 12.632 meter high.