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an air pump does 5,600 J or 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 degrees to the ground, what is the horizontal distance the water bottle rocket travels? round your answer to two significant figures.

2 Answers

7 votes
The answer is 5.3*10^1
User Gregoryp
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8.4k points
3 votes

As we know that the work done is given as


W = Fdcos\theta

here we know that

W = 5600 J

F = 150 N


\theta = 45 degree

now we will have


5600 = 150 * d * cos45

Now solving for distance "d"


d = (5600)/(150* cos45)


d = 5.3 * 10^1 m

so above is the horizontal distance moved

User AkiRoss
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7.8k points