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What is the maximum model rocket altitude?

  • Thread starter Paymemoney
  • Start date
  • #1
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Hi
I have a few question i don't understand how to complete.

Homework Statement


A model rocket leaves the ground, heading straight up at 49m/s.
a)What is the maximum altitude?What are its speed and altitude at b)1s c)4s

2. Homework Equations ( For all questions)
Constant acceleration formulas
Average speed equation

The Attempt at a Solution


Question a) i understand how to complete, however how would i solve b)

This is what i have tried:

t=1s
v-final=?
v-initial=0
a=9.80

used the v-final=v-initial + at
v-final = 9.8m/s

this is incorrect what have i done wrong?


Homework Statement


A jetliner leaves San Francisco for New York, 4600km away. With a strong tailwind, its speed is 110km/h. At the same time, a second jet leaves New York for San Francisco. Flying into the wind, it makes only 700km/h. When and where do the two planes pass each other?

The Attempt at a Solution


Well firstly i tried to find the time from NY to SF and NY to SF.

So i get 4.18hrs and 6.57hrs. I minus the figures to get the time that the two planes pass which is 2.37hrs. However according to the book's answer this is incorrect.

For the distance i went 400(1100-700) * 2.39 = 956km.

Homework Statement


A hockey puck moving at 32m/s slams through a wall of snow 35cm thick. It emerges moving at 18m/s. a) How much time does it spend in the snow? b) How thick a wall of snow would be needed to stop the puck entirely.

The Attempt at a Solution



i found question a) was t=0.014s, but could not find the correct answer to question b)

I have tried doing the following:

[tex]x= \frac{v-initial - v-final}{2} * t[/tex]

[tex]x=\frac{32}{2} * 0.014[/tex]

x=0.224m

P.S
 

Answers and Replies

  • #2
286
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I will answer your first question..

in the question it is given that v-initial = 49m/s , so why in part(b) you said it is zero??
another thing, when you are considering such motion , you will need the following equations:

v-final = v-initial - gt , (notice the minus sign befor g, since the gravity always downward)

v-final^2 = v-initial^2 - 2gy (y is the height)

y = v-inital*t - 0.5gt^2

try again and tell us what you get ..
 
  • #3
175
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ok, i got the answers to the altitude but i cannot seem to get the right answer for speed.

For the first one with 1s i got 9.8m/s, answers says its incorrect.
 
  • #4
286
0
are you sure that you substitue in with the right given value!!
please show me what equation you used , and what values you substitute there?..
 
  • #5
175
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after thinking about it i have got the answer
i used

v-final = v-initial + at

where v-final is what i am trying to find.

so it would be: 49-9.8 which is 39.2m/s

before i made v-initial 0m/s which it fact it is 49m/s.
 
  • #6
286
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exactly :) .. maybe you didnt notice that i mentioned in my first post that in the question v-initial is 49m/s ..
 
  • #7
ideasrule
Homework Helper
2,266
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I have tried doing the following:

[tex]x= \frac{v-initial - v-final}{2} * t[/tex]

[tex]x=\frac{32}{2} * 0.014[/tex]

x=0.224m

P.S
t isn't 0.014 anymore because the puck has to spend a longer time in the snow. Try calculating the acceleration of the puck, then using that to find the stopping distance.
 
  • #8
175
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t isn't 0.014 anymore because the puck has to spend a longer time in the snow. Try calculating the acceleration of the puck, then using that to find the stopping distance.
so what would the time be, i got the v-initial=32m/s, v-final=0m/s and i need at least three values to find acceleration.
 
  • #9
rl.bhat
Homework Helper
4,433
7
so what would the time be, i got the v-initial=32m/s, v-final=0m/s and i need at least three values to find acceleration.
Using the equation
vf^2 = vi^2 + 2*a*s, find the acceleration of the puck in the snow. Then you can find the time which the puck takes to cross the ice.
Using this value of acceleration, and taking the final velocity as zero, you can find the thickness of the ice to stop the puck.
 
  • #10
175
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Using the equation
vf^2 = vi^2 + 2*a*s, find the acceleration of the puck in the snow. Then you can find the time which the puck takes to cross the ice.
Using this value of acceleration, and taking the final velocity as zero, you can find the thickness of the ice to stop the puck.
yeh, but to find this i need at least three values. i only got 2 how can i find the acceleration?:

v-final=0m/s
v-initial=32m/s
a=?
t=?
x=?

I can't define x(s) as some given value, because that is what i am eventually going to find.
 
  • #11
286
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for question(3) part(a) .. you are given v-initial 32m/s , v-final 18m/s and x = 0.35m to find the time:


1) use this equation (v-final^2=v-initial^2 + 2ax) to find a ..
2) then use (v-final = v-initial + at ) to find the time ..


for part(b) .. you are given v-initial 32m/s , v-final 0m/s , and you are asked for x=?

1) since you got the acceleration a from part(a) use it along with the equation (v-final^2=v-initial^2 + 2ax) to find x ..



hopefully this is clear enough :) .. try now once more ..
 
  • #12
175
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for question(3) part(a) .. you are given v-initial 32m/s , v-final 18m/s and x = 0.35m to find the time:


1) use this equation (v-final^2=v-initial^2 + 2ax) to find a ..
2) then use (v-final = v-initial + at ) to find the time ..


for part(b) .. you are given v-initial 32m/s , v-final 0m/s , and you are asked for x=?

1) since you got the acceleration a from part(a) use it along with the equation (v-final^2=v-initial^2 + 2ax) to find x ..



hopefully this is clear enough :) .. try now once more ..
yep, that makes more sense and i have got the answer to be 0.512m
 
  • #13
286
0
I believe you got the right answer now :) since obviously x in part(b) should be greater than 0.35 ..
 
  • #14
175
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so can anyone help me on my second question about the jetliner?
 
  • #15
rl.bhat
Homework Helper
4,433
7
so can anyone help me on my second question about the jetliner?
If they meet after time t at a distance x from New york, then
First jet travels a distance x km and the other jet travels (4600 - x). Velocities of jet is known. Find t and equate them. Then solve for x.
 
  • #16
175
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so would i use the v=d/t formula or the constant acceleration formula.
Because when i used the v=d/t formula i didn't get the correct answer.
 
  • #17
rl.bhat
Homework Helper
4,433
7
so would i use the v=d/t formula or the constant acceleration formula.
Because when i used the v=d/t formula i didn't get the correct answer.
The jets are not accelerating.
You have to write
t = x/v1 = (4600 - x)/v2.
Solve the equation to find x.
 
  • #18
175
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how can i find 't' if i only have the values of velocity?

If i used the total distance of 4600/110(v1), would this be correct?
 
  • #19
rl.bhat
Homework Helper
4,433
7
You know the values of v1 and v2. Substitute them in
x/v1 = (4600 - x)/v2, and solve for x.
 
  • #20
175
0
ok, this is what i have done:

305.56(4600-x) = 194.4(x)
1405576-305.56x = 194.4(x)
499.96x = 1405576
x = 2811.38m

book's answers is 1800km what have i done wrong?
 
  • #21
rl.bhat
Homework Helper
4,433
7
ok, this is what i have done:

305.56(4600-x) = 194.4(x)
1405576-305.56x = 194.4(x)
499.96x = 1405576
x = 2811.38m

book's answers is 1800km what have i done wrong?
From one end x is nearly 2800 km. From other end the distance is
4600 - 2800 = 1800 km.
 

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