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

The cliff is 12.632 meter high.

Step-by-step 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 +(1)/(2)\cdot g\cdot t^(2)` (1)

Second stone

`y_(2) = y_(o,2) +v_(o,2)\cdot (t-2) +(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\,(m)/(s)`,
`v_(o,2) = -30\,(m)/(s)` and
`g = -9.807\,(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 +(1)/(2)\cdot g\cdot t^(2)`

`y_(o,1) = 12.632\,m`

The cliff is 12.632 meter high.