Answer:
The speed of the stone when it is 4.66 m higher is 236.057 m/s.
Explanation:
Given the initial velocity and vertical distance, we can use the fourth kinematic equation ([tex]v^{2} =v_{o}^{2}+2ay[/tex]) to find v final, or the v to the left of the equal sign. We know [tex]v_{o}[/tex] (initial velocity) is 24.7 m/s, y (change in vertical distance) is 4.66 m, and a is another way to write g (acceleration due to gravity), or 9.8 [tex]m/s^{2}[/tex].
From here you could plug in the values and solve for v final, but to make the solving process simpler, we can simplify the given equation, then plug in the known values.
To isolate v final, we can take the square root of [tex]v^{2}[/tex] and do the same to the right side of the equation. Therefore, we can find v final with: [tex]v_{o} \sqrt{2ay}[/tex], where v initial is outside of the square root because it squared...
If we plug in the known values to the simplified equation, we get: [tex]v=24.7m/s*\sqrt{2(9.8m/s)(4.66 m)}[/tex]
The final answer is 236.057 m/s.
