Question

A stone is thrown vertically upward with an initial speed of 24.7 m/s. Neglect air resistance. The speed of the stone when it is 4.66 m higher is:

Answer 1

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 ($v^{2} =v_{o}^{2}+2ay$) to find v final, or the v to the left of the equal sign. We know $v_{o}$ (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 $m/s^{2}$.

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 $v^{2}$ and do the same to the right side of the equation. Therefore, we can find v final with: $v_{o} \sqrt{2ay}$, 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: $v=24.7m/s*\sqrt{2(9.8m/s)(4.66 m)}$

The final answer is 236.057 m/s.