What is the Maximum Altitude?

  • #1

Homework Statement


A rocket, initially at rest, is fired vertically with an upward aceleration of 10m/s². At an altitude of .50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?


Homework Equations


V² = V_o² + 2aΔx
t = (V-V_o)/a

The Attempt at a Solution


Well I wanted to find the Final Velocity at the .50km first so

V² = V_o² + 2aΔx
V² = 0² + 2(10)(500m) *Converted .50km to m
V² = 10000
V = 100m/s

Since that final velocity

t = (V-V_o)/a
t = (100 - 0) / 10
t = 10seconds

So it takes 10 seconds to travel the 500m

But now I'm stuck...I see that to find the final altitude:

The X_o is now 500m
The V_o is now 100m/s

I believe that gravity -9.81m/s/s would become the new acceleration?

I'm not sure how to find out how much longer it will be in the air until it starts going down and I'm not sure how to find the final altitude.

Is this correct to this point?

EDIT: Actually the V would be 0m/s wouldn't it? I forgot that.

Now shouldn't I be able to

V = V_o + at
t = (0-100)/(-9.81)
t = 10.19367992 seconds remaining in the air so

X = V_o t + X_o + 1/2 a t²
X = 100(10.19367992) + 500 + (1/2)(-9.81)(10.19367992)²
X = 101.9367992 + 500 + (-509.6839961)
X = 92.2528031 m more than 500

So 592.2528031m

or

.59 km? No that can't be right...Okay now any help?
 
Last edited:

Answers and Replies

  • #2
You're very close. I think you just made a very simple arithmetic mistake: 100(10.19367992) ≠ 101.9367992
 
  • #3
You're very close. I think you just made a very simple arithmetic mistake: 100(10.19367992) ≠ 101.9367992

Ahh, didn't even notice that
Okay

X = V_o t + X_o + 1/2 a t²
X = 100(10.19367992) + 500 + (1/2)(-9.81)(10.19367992)²
X = 1019.367992 + 500 + (-509.6839961)
X = 1009.683996 m more than 500

So now

1009.683996 + 500 = 1509.683996m

Answer in km would be 1509.683996m / 1000 = 1.509684 km?

Look correct now?

Odd because I see the answer would be 1.0km

But i'm sure somewhere along the way there was another minor problem...like not carrying over a significant figure or something
 
  • #4
X = 1019.367992 + 500 + (-509.6839961)
X = 1009.683996 m more than 500

Again, very close ;) You already added the 500 m from when the engine cut off so your final answer would be 1009 m ≈ 1.0 km.
 
  • #5
Again, very close ;) You already added the 500 m from when the engine cut off so your final answer would be 1009 m ≈ 1.0 km.

Oh ARGH! I hate making those kinds of mistakes!! haha Thank you Melon...I really need to get back in the physics mindframe. Did awesome taking AP classes in high school but now that I'm in college it's like I cant get used to it again.
 

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