Answer:
[tex]v_2=\sqrt{2}v_1[/tex]
Explanation:
The velocity v₁ can be calculated with the kinematic formula:
[tex]v_1^{2} =v_0^{2} +2gh[/tex]
Since the object is initially at rest, v₁ becomes:
[tex]v_1=\sqrt{2gh}[/tex]
Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:
[tex]v_2^{2}=v_1^{2} +2gh[/tex]
Substituting v₁ in this expression and solving for v₂, we get:
[tex]v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}[/tex]
Now, dividing v₂ over v₁, we get the expression:
[tex]\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1[/tex]
It means that v₂ is √2 times v₁.
