Maximum height of a toy rocket?

  • #1
maximum height of a toy rocket??

Homework Statement


A toy rocket is launched vertically from ground level(y=0) at time t=0s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and aquired a velocity of 60m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The maximum height reached by the rocket is closest to
A. 244m
B. 256m-wrong
C. 221m
D. 233m
E. 209m


Homework Equations



y=vy*t+1/2g*(t^2)

The Attempt at a Solution


0= 60m/s*t+4.9m/s^2(t^2)
yaya not sure what i just did.
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,171
509


Try another of the kinematic eqautions, knowing that when the rocket reaches the top of its flight, its motion comes to a temporary stop.
 
  • #3


would it be something like vf^2=v0^2+2g(change in)y ??
 
  • #4
I like Serena
Homework Helper
6,579
176


would it be something like vf^2=v0^2+2g(change in)y ??

Close! :smile:
Actually vf^2=v0^2-2g(change in)y
since the acceleration is -g.
 
  • #5


okay but now i forgot what i need to plug in where. i know vf is 60 and i know g but am i looking for y or v0?
 
  • #6
87
0


I would solve it by finding how high the rocket went with the constant acceleration and then the velocity at the time that the acceleration stops. Then I would put the final velocity as the initial velocity for the second equation of displacement formula. Take the derivative of that and solve for time when the velocity is zero (at its highest point). Then I would think about what left I have to do to solve for the total displacement traveled at its highest point.
 
  • #7


okay you just lost me. too much at once.
 
  • #8
I like Serena
Homework Helper
6,579
176


okay but now i forgot what i need to plug in where. i know vf is 60 and i know g but am i looking for y or v0?

You need to split the trajectory into separate parts.
First part is when the engine is burning.
The problem states it gets up to a height of 49 meters, and at that point it has a velocity of 60 m/s.

Then you get the second part of the trajectory where you need to use the formula.
At that time the "initial velocity" is 60, and the trajectory continues until the rocket is at its highest point.
What is the "change in y" for this second part of the trajectory?
 
  • #11
87
0


The answer I got was closest to D. 233m. That is correct in your textbook?
 
  • #12


umm i did 60^2 + (9.8*2)
 
  • #13


The answer I got was closest to D. 233m. That is correct in your textbook?

this text book doesnt have answers. how did you get that??
 
  • #14
87
0


I am not sure if I can show the step-by-step mathematics here to give you the answer. It involves basic calculus and rearranging the formulas of the the position function. A major hint is this equation:

Let up be positive:

x(t) = -1/2gt^2 + 60t, where g = 9.8 m/s/s
 
  • #15
I like Serena
Homework Helper
6,579
176


umm i did 60^2 + (9.8*2)

I'm afraid that is not right.

Your formula is:
[tex]v_f^2=v_0^2-2g\Delta y[/tex]
To find (change in)y, you'd have:
[tex]\Delta y = {v_0^2 - v_f^2 \over 2g}[/tex]
 
  • #16


yeah i figured it wasn't. i followed the formula you just game me and i came up with change in y= 2.5m is this right?
 
  • #17
87
0


yeah i figured it wasn't. i followed the formula you just game me and i came up with change in y= 2.5m is this right?

What person are you talking to? :tongue:
 
  • #18


What person are you talking to? :tongue:

i was talking to i like serena, his equation was easier to figure out than yours was.
 
  • #19
I like Serena
Homework Helper
6,579
176


yeah i figured it wasn't. i followed the formula you just game me and i came up with change in y= 2.5m is this right?

Noooooooo. :wink:

Try:
[tex]\Delta y = {v_0^2 - v_f^2 \over 2g} = {60^2 - 0^2 \over 2 \cdot 9.8}[/tex]
 
  • #20
87
0


My equation is the position traveled when gravity takes over. It is simplified to x(t) = -4.9t^2 + 60t
 
  • #21


Noooooooo. :wink:

Try:
[tex]\Delta y = {v_0^2 - v_f^2 \over 2g} = {60^2 - 0^2 \over 2 \cdot 9.8}[/tex]

haha i think i forgot to sqaure 60. now i got 183.67
 
  • #22
87
0


haha i think i forgot to sqaure 60. now i got 183.67

Great! Now combine it to the first amount of distance traveled to get the highest point.
 
  • #23
I like Serena
Homework Helper
6,579
176


haha i think i forgot to sqaure 60. now i got 183.67

Good! :smile:

Now add the height that was achieved in the first part of the trajectory?
 
  • #24


232.67!!! thats close to 233. yay we did it!
 

Related Threads on Maximum height of a toy rocket?

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
10
Views
2K
Replies
1
Views
3K
Replies
15
Views
1K
Replies
1
Views
2K
Replies
2
Views
13K
Replies
3
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
4
Views
6K
Top