Rocket Height-Kinematic Question

[Chandasouk]

Homework Statement

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t1, when the fuel is exhausted.

Find the maximum height H that the rocket reaches (neglecting air resistance).

Express the maximum height in terms of a, t1, and/or g. Note that in this problem, g is a positive number equal to the magnitude of the acceleration due to gravity.

The Attempt at a Solution

Okay, in this problem, i am pretty much stuck. I only did the hints that Mastering Physics suggested I do to help me solve this problem.

I found the height, H1, above the ground at which the rocket exhausted its fuel

H1=1/2at1^2

I then found the y velocity, Vy1, that the rocket has when the engine runs out of fuel

Vy1=at1

Where do I go from here?

 

[pgardn]

So you got the initial velocity when the rocket runs out of fuel (stops accelerating up) so it is now in free fall. What is the velocity of the rocket when it reaches its max height? Then use the kinematic equation that does not involve time but has displacement in the equation. Essentially the last part of the trip is just like throwing a rock up in the air. You got the velocity upon release, and at max height the velocity is zero.

 

[Doc Al]

Once the rocket runs out of fuel it can be treated like any other projectile in freefall. You have its speed when it runs out of fuel, so figure out how much further it rises.

 

[Chandasouk]

Okay, the equation which pgardn suggested is

Vf^2 = Vi^2 + 2ad


The velocity of the rocket at the max height of it's path is 0m/s

Acceleration during free fall equals -9.81m/s^2

I am assuming the Final Velocity (Vf) is 0m/s, so I set it up like this

(0m/s)^2 = Vi^2 + 2ad

Previously, I found the Y velocity which was Vy1 = at1. I use this as a substitute


(0m/s)^2 = (at1)^2 + 2(-9.81m/s^2)d

0-(at1)^2=-19.62m/s^2*d

-(at1)^2
------------- = D ?
-19.62m/s^2


In terms of a, t1, and/or g, it would be

H = -(at)^2
-------- ?
- 2g

 

[Doc Al]

In terms of a, t1, and/or g, it would be

H = -(at)^2
-------- ?
- 2g

That's the additional height the rocket will reach measured from the point where it ran out of fuel. What's the total height above the ground? (And please cancel those minus signs.)

 

[Chandasouk]

So, previously, I found the height, H1, above the ground at which the rocket exhausted its fuel

H1=1/2at1^2

and also using my last equation

H = (at)^2
--------
2g
The total distance covered by the rocket would equal the sum of these two equations

H=1/2at1^2 + (at1^2/2g) ?

 

[Doc Al]

The total distance covered by the rocket would equal the sum of these two equations

H=1/2at1^2 + (at1^2/2g) ?

Looks good to me.

 

[Chandasouk]

Mastering Physics told me this answer was incorrect because

One of your terms has dimensions of time squared, not length.

 

[rl.bhat]

So, previously, I found the height, H1, above the ground at which the rocket exhausted its fuel

H1=1/2at1^2

and also using my last equation

H = (at)^2
--------
2g
The total distance covered by the rocket would equal the sum of these two equations

H=1/2at1^2 + (at1^2/2g) ?

It should be
H = 1/2*at_1² +1/2 (at₂)²/g

 

[Doc Al]

The total distance covered by the rocket would equal the sum of these two equations

H=1/2at1^2 + (at1^2/2g) ?

You misplaced your parentheses. It should be:
H=1/2at1^2 + (at1)^2/2g

(Sorry for not catching that earlier.)

Use subscripts/superscripts for further clarity:
H=1/2at₁² + (at₁)²/2g