Question

an air pump does 5,600 j of work to launch a water bottle rocket into the air. if the air pump applies 150 n of force to the rocket at an angle of 45° to the ground, what is the horizontal distance the water bottle rocket travels? round your answer to two significant figures.

• ( 17 \times 10^1 , \text{m} )
• ( 5.3 \times 10^1 , \text{m} )
• ( 1.1 \times 10^2 , \text{m} )
• ( 5.9 \times 10^2 , \text{m} )

Explanation:

Step1: Recall the work formula

The formula for work done by a force is \(W = Fd\cos\theta\), where \(W\) is work, \(F\) is force, \(d\) is distance, and \(\theta\) is the angle between the force and the displacement. We need to find the horizontal distance \(d\), so we can rearrange the formula to \(d=\frac{W}{F\cos\theta}\).

Step2: Identify the given values

We know that \(W = 5600\space J\), \(F = 150\space N\), and \(\theta = 45^{\circ}\). The cosine of \(45^{\circ}\) is \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.7071\).

Step3: Substitute the values into the formula

Substitute \(W = 5600\), \(F = 150\), and \(\cos\theta\approx0.7071\) into the formula \(d=\frac{W}{F\cos\theta}\):
\[

$$\begin{align*} d&=\frac{5600}{150\times0.7071}\\ &=\frac{5600}{106.065}\\ &\approx52.79\\ &\approx5.3\times 10^{1}\space m \end{align*}$$

\]

Answer:

\(5.3 \times 10^{1}\space m\)