To find the horizontal distance the water bottle rocket travels, we need to use the work-energy principle. The work done on an object can be described by the formula:
[tex]\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \][/tex]
Where:
- [tex]\(\text{Work}\)[/tex] is the work done, which is 5600 J (Joules).
- [tex]\(\text{Force}\)[/tex] is the applied force, which is 150 N (Newtons).
- [tex]\(\theta\)[/tex] is the angle of force application relative to the ground, which is [tex]\(45^\circ\)[/tex].
- [tex]\(\text{Distance}\)[/tex] is what we aim to find, specifically the horizontal distance.
First, we resolve the force applied into its horizontal component. The horizontal component of the applied force can be found using trigonometry and specifically the cosine function, as we know:
[tex]\[ \cos(45^\circ) = \frac{\sqrt{2}}{2} \][/tex]
Thus, the formula for the horizontal distance can be rearranged to solve for [tex]\(\text{Distance}\)[/tex]:
[tex]\[ \text{Distance} = \frac{\text{Work}}{\text{Force} \times \cos(\theta)} \][/tex]
Let's plug in the values:
[tex]\[ \text{Distance} = \frac{5600 \, \text{J}}{150 \, \text{N} \times \cos(45^\circ)} \][/tex]
Convert the angle [tex]\(45^\circ\)[/tex] to radians for calculation purposes:
[tex]\[ \cos(45^\circ) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \][/tex]
Now we can express the distance:
[tex]\[ \text{Distance} = \frac{5600}{150 \times \frac{\sqrt{2}}{2}} \][/tex]
Simplifying the denominator:
[tex]\[ \text{Distance} = \frac{5600}{150 \times 0.707} \approx \frac{5600}{106.05} \approx 52.797306328595546 \][/tex]
Rounding this value to two significant figures:
[tex]\[ \text{Distance rounded} = 52.8 \, \text{meters} \][/tex]
So, the correct horizontal distance the water bottle rocket travels is:
[tex]\[ \boxed{5.3 \times 10^1 \, \text{m}} \][/tex]
