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### Kinematics - Finding Rocket Height and Speed

```
Date: 10/15/2001 at 17:44:01
From: Chuck Baker
Subject: Kinematics

During a test a rocket is traveling upward at 75m/s, and when it is
40 m from the ground its engine fails. I would like to determine the
maximum height, h, reached by the rocket and its speed just before it
hits the ground. While in motion, the rocket is subjected to a
constant downward acceleration of 9.81 m/s squared due to gravity.

In this case, I used the known force on the rocket and related that
to its acceleration using the equation of motion. I then used
Kinematics to relate the projectile's acceleration to its position.

I arrived at an answer of h = 427 meters and the impact velocity
of 92.1 m/s

Can someone check my methodology and results?
```

```
Date: 10/16/2001 at 11:32:02
From: Doctor Code
Subject: Re: Kinematics

Hi Chuck,

For this problem we're assuming that the mass of the rocket doesn't
change; in other words, it's not burning any propellant. If it did,
you'd have to use the rocket equation, but since we're ignoring that,
you can get away with using only ballistic equations.

If I'm understanding the problem correctly, during the first 40 meters
of ascent, the rocket travels at a constant velocity of 75 m/s,
because the engine thrust exactly balances the force of gravity,
resulting in no net acceleration. But at 40 meters, the engine fails,
and at this point, the rocket becomes a pure ballistic object subject
only to gravitational acceleration.

In ballistics, you assume constant acceleration:

a = g

where g = -9.81 m/s^2. Integrating with respect to time gives an
equation for velocity:

vf = g * t + vi

where t is time, vf is the final velocity and vi is the initial
velocity. Integrate again to get an equation for position:

yf = (1/2) * g * t^2 + vo * t + yi

where yf is the final position and yi is the initial position.  If we
solve the velocity equation for time:

t = (vf - vi) / g

But we know that the final velocity of the rocket is 0:

t = -vi / g

Now plug in this value for time into the equation for position:

vi^2
yf = ------ + yi
-2*g

Plugging in 75 m/s for vi, 40 m for yi, and -9.81 m/s^2 for g gives
yf = 326.7 meters apex height.

Write back if you need more help.

- Doctor Code, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/16/2001 at 11:56:55
From: Baker, Charles
Subject: Re: Kinematics

Hello Dr. Math,

I understand your results thanks, but how do you determine the speed
of the rocket before it hits the ground? Impact velocity?

Thanks very much,
Chuck
```

```
Date: 10/22/2001 at 09:21:46
From: Doctor Code
Subject: Re: Kinematics

Hi Chuck,

There are two phases for the ascent. The first phase, up to 40 meters,
is non-ballistic, because the rocket is travelling at constant
velocity. The second phase, from 40 to 327 meters, is ballistic,
because the rocket is only subject to gravity, no engine. The descent
is completely ballistic.

In order to simplify our lives, we can pretend that the ascent of the
rocket was actually completely ballistic. In other words, the rocket
leaps into the air with some initial velocity, and the engine cuts out
immediately, so that the rocket is only affected by gravity. This
helps us because ballistic paths are symmetric (parabolas or vertical
lines).

Let's figure out at what velocity the rocket would need in order to
reach a height of 326.7 meters.

This can be found by rearranging my original equation:

vi^2
yf = ------ + yi
-2*g

vi = sqrt((yf - yi) * -2 * g)

= sqrt((327 - 0)* -2 * -9.81)

= 80 m/s

At the apex, we're still at a height of 326.7 meters, and 0 velocity,
it's just that we got there in a different way.

Since we took off with a speed of 80 m/s, and travelled up to 326.7
meters, if we fall back down to the ground, the impact velocity will
also be 80 m/s.

Write back if you need more help.

- Doctor Code, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Physics/Chemistry

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