Question

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.

Answer 1

The answer is 5.3*10^1

Answer 2

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

$5600 = 150 \times d \times cos45$

Now solving for distance "d"

$d = \frac{5600}{150\times cos45}$

$d = 5.3 \times 10^1 m$

so above is the horizontal distance moved