Question

an object is in free fall for 10.00 seconds. what was the velocity at the end of the time interval? (be sure you use the correct number of significant figures) (include unit: m, s, m/s, m/s2) (put a space in between number and unit, ex. 10 m/s)

Explanation:

Step1: Identify known values

In free fall, initial vertical velocity \( \vec{v}_{0y} = 0 \, \text{m/s} \) (assuming dropped from rest), acceleration \( \vec{a}_y = g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), time \( \Delta t = 10.00 \, \text{s} \).

Step2: Substitute into velocity formula

Use \( \vec{v}_{fy} = \vec{v}_{0y} + \vec{a}_y \Delta t \). Substitute \( \vec{v}_{0y} = 0 \, \text{m/s} \), \( \vec{a}_y = 9.8 \, \text{m/s}^2 \), \( \Delta t = 10.00 \, \text{s} \):
\( \vec{v}_{fy} = 0 + 9.8 \times 10.00 \)
\( \vec{v}_{fy} = 98.0 \, \text{m/s} \) (considering significant figures, 10.00 has 4, 9.8 has 2, but usually \( g = 9.81 \, \text{m/s}^2 \) for more precision; if using \( 9.81 \)):
\( \vec{v}_{fy} = 0 + 9.81 \times 10.00 = 98.1 \, \text{m/s} \). But common approximation is \( 9.8 \), so \( 98.0 \) or \( 98 \) (depending on \( g \) used).

Answer:

98.0 m/s (or 98 m/s, or 98.1 m/s if using \( g = 9.81 \, \text{m/s}^2 \))