Question

A toy rocket is launched vertically from ground level (y = 0 m), at time t=0 s. The rocket engine provides constant upward acceleration during the burning phase. At the instant of engine burnout, the rocket has risen to 68 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground.

Answer 1

1.The time interval, during which the rocket engine provides upward acceleration, is closest to
5.0 s
3.0 s
4.0 s
4.5 s
3.5 s
2.The upward acceleration, in m/s2, of the rocket during the burn phase is closest to
6.0
5.8
6.4
6.2
6.6
3.The maximum height, in meters, reached by the rocket is closest to
125
105
115
120
110
4.The time interval, during which the rocket is in unpowered flight, is closest to
7.0 s
7.5 s
8.0 s
6.5 s
6.0 s
5.The speed of the rocket, in m/s, upon impact on the ground is closest to
37
52
47
57
42

Answer 2

So, you have \(y_0\), \(v_0\), \(t_0\), and the \(y\) and \(v\) at which the rocket stops accelerating. Do you have an equation in mind?

Answer 3

wait its not vf^2=vi^2+2(a)(d) is it?

Answer 4

It definitely is!!

Answer 5

Wait, not that one.. There's no time, t!
One sec.

Answer 6

\[x−x_0 =\frac{v_0 +v_2}{ 2}\ t\] from @Festinger

Answer 7

whats x= again?

Answer 8

Make that a \(y\), since we're dealing with the vertical :)

Answer 9

\(y\) is the height at the end of the interval.

Answer 10

so 68-0= (0+30/2)(t)
which ---> 4.53=t

Answer 11

is that right?

Answer 12

ok can you help me with the next part of the question

Answer 13

... and how do i differentiate between m/s and m/s2

Answer 14

That's what I got! So now you have \(\Delta t\), the time elapsed while the rocket was accelerating upwards. There's a handy equation that is pretty much the definition of acceleration put into math. Do you know it?

Answer 15

It was actually equation (2) from Festinger.
\[a=\frac{\Delta v}{\Delta t}\]

Answer 16

haha i actually knew that one!

Answer 17

so 30-0/4.5-0
which ---> a=6.6
is that right?

Answer 18

ok next part

Answer 19

Again, got the same thing! Sorry, I'm multitasking.
For this one did you have an equation in mind?

Answer 20

x=x0+v0t+1/2at^2

x−x0=v0+v2t

Answer 21

one of those right?

Answer 22

And m/s is velocity, while \(m/s^2\) is acceleration.

You have to make sure you're thinking of the right variables for this new interval, which is after the rocket boosted up. Now you don't necessarily know the time interval until it reaches the top. But you do know it stops at the top before heading down, so the final velocity is 0.

To start, you know you're up 68m, and you have a velocity of \(30\ m/s\) to start with and \(0\ m/s\) at the end. You use \(-9.8\ m/s^2\) for gravitational acceleration, right?

Answer 23

Or \(-10\ m/s^2\)?

Answer 24

right

Answer 25

For this second interval, you can use \(v^2 = v_0^2+2ad\) to find the distance while the rocket was not boosting. Then you can add it to the 68m you were up already.

Answer 26

i got 45.91+68 to get 114
so about 115 correct?

Answer 27

Yeah! I got 45 when using \(10\ m/s^2\), and same as you with \(9.8\ m/s^2\), so yeah!

Answer 28

perfect! how bout the next question

Answer 29

Well, the first part of unpowered flight is what we just worked with. The initial \(v\) was \(30\ m/s\), the final \(v\) was \(0\). The acceleration was \(-10\ m/s\), so you can use the definition of acceleration again!

Answer 30

but that didnt work... i got 3, wouldnt i need at east one time (initial or final)

Answer 31

what did you get?

Answer 32

Yeah, that interval is about 3s. But then there's the entire fall to go yet! You know how high you are (\(d\)), and the acceleration from gravity (\(-10\ m/s^2\)), and the velocity you start with at the top. You need the time!

Answer 33

I sort of want to find #5 first... To get the final velocity. But there's probably an equation to go on with #4!

Answer 34

lets do number 5 first then

Answer 35

I agree.

Answer 36

i honestly have no clue how you do number 4 or 5

Answer 37

Haha, #5, then.

Answer 38

So, let me draw a picture. They usually help. It'll be sloppy, but quick. I'll draw only the fall.

Answer 39

\[v^2 = v_0^2+2ad\]

Answer 40

whats d=

Answer 41

115 or 68?

Answer 42

115 right?

Answer 43

115, yeah!

Answer 44

That was our max height...

Answer 45

would we want to use 9.8 sense it is asking in m/s

Answer 46

\[v = \sqrt{v_0^2+2ad}\]

Answer 47

\(a\) is about \(-10\ m/s^2\), and \(v\) will come out to be in \(m/s\).

Answer 48

ok

Answer 49

i got 47.95 so about 47 right?

Answer 50

I got the same! Back to #4!

Answer 51

We found the freefall going up to be 3s.
Now we have...And the handy definition of acceleration says that the time for the descent is...

Answer 52

\[a=\frac {v-v_0}{\Delta t}\]

Answer 53

i got 4.7 thats not rght

Answer 54

But that was only the freefall descent. There was also the freefall ascent. You see where I'm going?

Answer 55

you add them? idk

Answer 56

Take a guess.

Answer 57

is it 7.5?

Answer 58

That's what I got at first, but it was sort of between answers. So I used the real values. The time up was 30/9.8 and the time down was about 48/9.8, and I added them to get about 8s.

Answer 59

So, I think that's the answer it should want...

Answer 60

yea great thank you so much! i really appreciate it, and im glad you actually taught me how to do it instead of just giving me the answers... haave a good night and maybe you can help me again in the future. Thanks!

Answer 61

It's about 7.75 with \(a=10\ m/s^2\) though....

Answer 62

8 worked perfectly

Answer 63

I'd like to help if possible! And I wish those answers weren't so close together! Did you get them all right?

Answer 64

yup! 100% (it looks like there werent any point deductions for late answers either :) )

Answer 65

im going now but thank you again!

Answer 66

You're very welcome! And you did well! :)

Answer 67

Glad it all worked out.