Question

George drops a stone from atop a cliff of 25m. How long does it take to hit the ocean below, what velocity is it travelling when it gets there?

Answer 1

Given data:

* The height of the stone is 25 m.

* The initial velocity of the stone is 0 m/s.

* The acceleration due to gravity of the stone is,

`g=9.8ms^(-2)`

Solution:

(a). By te kinematics equation, the distance traveled by the stone in terms of time is,

`h=ut+(1)/(2)gt^2`

where h is the height, u is the initial velocity, t is the time taken, and g is the acceleration due to gravity,

Substituting the known values,

`\begin{gathered} 25=0+(1)/(2)*9.8* t^2 \\ 25=4.9* t^2 \\ t^2=(25)/(4.9) \\ t^2=5.1 \\ t=2.26\text{ s} \end{gathered}`

Thus, the time taken by the stone to reach the ground is 2.26 seconds.

(b). By the kinematics equation, the final velocity of the stone in terms of the initial velocity is,

`v-u=gt`

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken,

Substituting the known values,

`\begin{gathered} v-0=9.8*2.26 \\ v=22.14ms^(-1) \end{gathered}`

Thus, the final velocity of the stone is 22.14 meter per second.