Answer:
The velocity before it hits the ground is:
[tex]v=28.3~\text{m/s}[/tex]
Explanation:
In the first situation, we have the initial vertical velocity to be: [tex]10~\text{m/s}[/tex]
And the final vertical velocity is: [tex]30~\text{m/s}[/tex]
The acceleration here will be the value of [tex]g[/tex]: [tex]9.81[/tex], since we assume only gravity to be acting on the rock while it's falling.
We will make use of this formula throughout the question:
[tex]v^{2}=u^{2}+2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Substitute the values we have:
[tex]30^{2}=10^{2}+2(9.81)s\\\\2(9.81)s=30^{2}-10^{2}\\\\19.62s=800\\\\s=\frac{800}{19.62}\\\\s=40.77~\text{m}[/tex]
In the second situation, there is no initial vertical velocity.
There is only a horizontal velocity: [tex]10~\text{m/s}[/tex]
So, the initial vertical velocity is: [tex]0~\text{m/s}[/tex]
[tex]a=9.81[/tex] and we already know the height of the cliff to be: [tex]40.77~\text{m}[/tex]
Substitute the values we have and calculate the velocity:
[tex]v^{2}=0^{2}+2(9.81)(40.77)\\\\v^{2}=800\\\\v=28.3~\text{m/s}[/tex]
