Certainly! Let's solve the problem step-by-step.
Given:
- The mass of the rock, [tex]\( m = 2.4 \)[/tex] kg.
- The acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s[tex]\(^2\)[/tex].
The table provided gives the mechanical energy and velocity for various trials. To determine the height above the ground, we need to use the relationship between mechanical energy, kinetic energy, and potential energy.
Mechanical Energy Formula:
[tex]\[
E_m = KE + PE
\][/tex]
Where:
- [tex]\( KE \)[/tex] is the kinetic energy.
- [tex]\( PE \)[/tex] is the potential energy.
For each trial:
[tex]\[
E_m = \frac{1}{2} m v^2 + mgh
\][/tex]
Rearranged to solve for height [tex]\( h \)[/tex]:
[tex]\[
h = \frac{E_m - \frac{1}{2} m v^2}{mg}
\][/tex]
Let's write down the mechanical energy [tex]\( E_m \)[/tex] and velocity [tex]\( v \)[/tex] for each trial, and calculate the height [tex]\( h \)[/tex]:
1. Trial 1:
[tex]\[
E_m = 176.4 \text{ J}, \quad v = 7.0 \text{ m/s}
\][/tex]
[tex]\[
h_1 = \frac{176.4 - \frac{1}{2} \cdot 2.4 \cdot (7.0)^2}{2.4 \cdot 9.8}
\][/tex]
[tex]\[
h_1 = \frac{176.4 - 58.8}{23.52} \approx 5.00 \text{ m}
\][/tex]
2. Trial 2:
[tex]\[
E_m = 157.7 \text{ J}, \quad v = 2.0 \text{ m/s}
\][/tex]
[tex]\[
h_2 = \frac{157.7 - \frac{1}{2} \cdot 2.4 \cdot (2.0)^2}{2.4 \cdot 9.8}
\][/tex]
[tex]\[
h_2 = \frac{157.7 - 4.8}{23.52} \approx 6.50 \text{ m}
\][/tex]
3. Trial 3:
[tex]\[
E_m = 170.2 \text{ J}, \quad v = 6.0 \text{ m/s}
\][/tex]
[tex]\[
h_3 = \frac{170.2 - \frac{1}{2} \cdot 2.4 \cdot (6.0)^2}{2.4 \cdot 9.8}
\][/tex]
[tex]\[
h_3 = \frac{170.2 - 43.2}{23.52} \approx 5.40 \text{ m}
\][/tex]
4. Trial 4:
[tex]\[
E_m = 123.7 \text{ J}, \quad v = 3.0 \text{ m/s}
\][/tex]
[tex]\[
h_4 = \frac{123.7 - \frac{1}{2} \cdot 2.4 \cdot (3.0)^2}{2.4 \cdot 9.8}
\][/tex]
[tex]\[
h_4 = \frac{123.7 - 10.8}{23.52} \approx 4.80 \text{ m}
\][/tex]
Now we have the heights:
- Trial 1: [tex]\( 5.00 \)[/tex] m
- Trial 2: [tex]\( 6.50 \)[/tex] m
- Trial 3: [tex]\( 5.40 \)[/tex] m
- Trial 4: [tex]\( 4.80 \)[/tex] m
The maximum height is reached during Trial 2, at [tex]\( 6.50 \)[/tex] m.
Thus, during Trial 2, the rock was highest above the ground.
