Question

draw a picture of the path
$v_x = \frac{\delta x_v}{t}$
$v_f^2 = v_i^2 + 2a \delta x$
$v_f^2 = 0^2 + 2(-9.81 m/s)(-62)$
$v_f^2 = 1216.44$
$-\sqrt{1216} = 34.86$

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \( v_f^2 = v_i^2 + 2a\Delta x \). Here, initial velocity \( v_i = 0 \, \text{m/s} \), acceleration \( a=- 9.81\,\text{m/s}^2 \), and displacement \( \Delta x=- 0.62\,\text{m} \) (assuming downward is negative, but let's check the signs). Wait, in the calculation, the user has \( 2(-9.81)(- 0.62) \)? Wait the written calculation is \( v_f^2=0^2 + 2(-9.81)(- 0.62) \)? Wait no, the hand - written is \( v_f^2 = 0^2+2(-9.81)(- 0.62) \)? Wait the user's calculation: \( v_f^2=0^2 + 2(-9.81)(- 0.62)=0 + 2\times9.81\times0.62=12.1644 \), then \( v_f=\sqrt{12.1644}\approx3.49\,\text{m/s} \) (but the user wrote \( -\sqrt{12.16} = 3.496 \), maybe a sign error, but let's follow the steps).

Step2: Substitute values into the equation

Given \( v_i = 0\,\text{m/s} \), \( a=-9.81\,\text{m/s}^2 \), \( \Delta x=- 0.62\,\text{m} \) (if we take upward as positive, and the object is moving downward, displacement is negative, acceleration is negative). Plug into \( v_f^2=v_i^2 + 2a\Delta x \):

\( v_f^2=0^2+2\times(-9.81)\times(- 0.62) \)

\( v_f^2 = 0+2\times9.81\times0.62 \)

\( v_f^2=12.1644 \)

Step3: Solve for \( v_f \)

Take the square root of both sides. Since velocity direction (if we consider the physical context, maybe downward, but the magnitude is what's important here). \( v_f=\sqrt{12.1644}\approx3.49\,\text{m/s} \) (the user's calculation has a small error in the square root, \( \sqrt{12.1644}\approx3.49 \), and if we consider the sign, maybe \( - 3.49\,\text{m/s} \) depending on the coordinate system, but the user wrote \( -\sqrt{12.16}=3.496 \), probably a sign mix - up, but the magnitude is approximately \( 3.5\,\text{m/s} \))

Answer:

The final velocity is approximately \( \boldsymbol{3.5\,\text{m/s}} \) (or \( - 3.5\,\text{m/s} \) depending on the coordinate system convention)