To determine which planet allows the space probe to achieve the highest speed after falling 50 meters, we will use the kinematic equation for velocity under constant acceleration:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Here,
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( u \)[/tex] is the initial velocity (which is 0 in this case since the probe starts from rest),
- [tex]\( a \)[/tex] is the acceleration due to gravity on the respective planet,
- [tex]\( s \)[/tex] is the distance fallen (50 meters).
Given that [tex]\( u = 0 \)[/tex], the equation simplifies to:
[tex]\[ v^2 = 2as \][/tex]
Thus,
[tex]\[ v = \sqrt{2as} \][/tex]
We will use the given gravitational accelerations for each planet to find the final velocities:
1. Venus:
- Acceleration due to gravity, [tex]\( a = 8.9 \, m/s^2 \)[/tex]
- Distance fallen, [tex]\( s = 50 \, m \)[/tex]
[tex]\[ v = \sqrt{2 \cdot 8.9 \cdot 50} \][/tex]
[tex]\[ v \approx 29.83 \, m/s \][/tex]
2. Earth:
- Acceleration due to gravity, [tex]\( a = 9.8 \, m/s^2 \)[/tex]
- Distance fallen, [tex]\( s = 50 \, m \)[/tex]
[tex]\[ v = \sqrt{2 \cdot 9.8 \cdot 50} \][/tex]
[tex]\[ v \approx 31.30 \, m/s \][/tex]
3. Uranus:
- Acceleration due to gravity, [tex]\( a = 8.7 \, m/s^2 \)[/tex]
- Distance fallen, [tex]\( s = 50 \, m \)[/tex]
[tex]\[ v = \sqrt{2 \cdot 8.7 \cdot 50} \][/tex]
[tex]\[ v \approx 29.50 \, m/s \][/tex]
4. Neptune:
- Acceleration due to gravity, [tex]\( a = 11.0 \, m/s^2 \)[/tex]
- Distance fallen, [tex]\( s = 50 \, m \)[/tex]
[tex]\[ v = \sqrt{2 \cdot 11.0 \cdot 50} \][/tex]
[tex]\[ v \approx 33.17 \, m/s \][/tex]
5. Saturn:
- Acceleration due to gravity, [tex]\( a = 9.0 \, m/s^2 \)[/tex]
- Distance fallen, [tex]\( s = 50 \, m \)[/tex]
[tex]\[ v = \sqrt{2 \cdot 9.0 \cdot 50} \][/tex]
[tex]\[ v \approx 30.00 \, m/s \][/tex]
By comparing the final velocities, we see that the highest final velocity is achieved on Neptune, at approximately [tex]\( 33.17 \, m/s \)[/tex].
Therefore, the correct answer is:
D. Neptune
