- #1

MonkeyDLuffy

- 5

- 0

## Homework Statement

A student experimenting with model rockets measures the speed of a vertically-launched rocket to be 18.0 m/s when it is 75.0 m above the ground on the way up. The rocket engine fires from when the rocket is at ground level to when it is 8.75 m above the ground. If the rocket has a mass of 0.893 kg, use the work-kinetic energy theorem to determine:

(a) The speed of the rocket when it was 8.75 m above the ground

(b) The maximum height attained by the rocket

(make sure you know what the work-kinetic energy theorem is before you start your solution; it does not, for example, use gravitational potential energy.)

## Homework Equations

1. ΔK = ½m(V

^{2}- V

_{0}

^{2})

2. W

_{net}= ΔK

3. W

_{gravity}= mgh

4. V = V

_{0}- gt

^{2}

5. y

_{max}= y

_{0}+ V

_{0}t - ½gt

^{2}

## The Attempt at a Solution

a. My first instinct was to calculate ΔK for the entire flight, which I found to be about 145 J. Though I am not entirely sure this is true because the engine stopped firing at 8.75 m. What pushed me to believe this was the ΔK ,and therefore the W

_{net}, of the flight up to y = 75.0 m was a quote from my book which reads, " the work-energy theorem is valid even without the assumption that the net force is constant." Unfortunately, I cannot say with certainty that this is even relevant to the problem. If anyone could clarify this I would be very thankful.

b. That seemed to be a dead end, so I calculated the work done by gravity at 8.75 m and assumed this was the net-work from y = 0 m to y = 8.75 m. But after re-reading the problem I found that to be foolish, and so I was back at square one.

c. I became a bit distressed and decided to hit point a again, I figured this:

If W

_{net}= 145 J at y = 75.0 m , then the engine must have done upward work from y = 0 m to y = 8.75 m such that,

W

_{engine}- W

_{gravity}(which is about 657 J at y = 75.0 m) = 145 J.

So W

_{engine}must be,

W

_{net}+ W

_{gravity}= 802 J.

If this were true, then:

W

_{engine}= ΔK

_{0→8.75m}

W

_{engine}= ½m(V

^{2}- 0) → √(2W

_{engine})/m = V

_{8.75m}→ V

_{8.75m}= 42.4 m/s

d. If this were all true, then the maximum height would be achieved when,

W

_{gravity}= W

_{engine}(a bit irrelevant but I'd like to know if this is true)

and I could use equation 4 above to solve for the time of the flight from y = 8.75 m to y

_{max}, which I could then put into equation 5 where y

_{0}= 8.75 m, and V

_{0}= 42.4 m/s

I hope at least one of the conclusions I drew will be of actual use (though I'm doubtful). Thank you in advance for pointing me in the right direction!