# Projectile motion question-nothing special.

1. Sep 26, 2011

Projectile motion question

1. The problem statement, all variables and given/known data

A 5200 kg cart carrying a vertical rocket launcher moves to the right at a constant speed of 33.0 m/s along a horizontal track. It launches a 49.4 kg rocket vertically upward with an initial speed of 43.1 m/s relative to the cart.

a)How high will the rocket go?

b)How far does the cart move while the rocket is in the air?

c)At what angle, relative to the horizontal, is the rocket traveling just as it leaves the cart, as measured by an observer at rest on the ground?

d) What is the rocket's trajectory as seen by an observer stationary on the cart.

the rocket travels in a cycloid
the rocket travels straight up and then straight down
the rocket travels in a parabola

e)What is the rocket's trajectory as seen by an observer stationary on the ground.
the rocket travels in a cycloid
the rocket travels straight up and then straight down
the rocket travels in a parabola
2. Relevant equations

Any constant acceleration equation.

Vo sin theta = Voy
Vo cos theta = Vox

3. The attempt at a solution

a) 0= Voy^2 + 2(-980)(Ymax)
ymax = 94.8 m

Is this right?

b) Then the total distance traveled by the cart, isn't it just 94.8 * 2 =189.6 m?

c) x sin theta = 43.1 m/s
x cos theta =38.0 m/s

theta = 48.59 degrees.

Is this right?

d) Is straight up and down correct, since Vox and Vx are constant with ax=o.

e)Is the parabola correct? I don't know why scientifically, but I'm pretty sure an observer(outside of cart) would see a parabola, since one obviously does occur.

Is this all right or not? Thanks.

Last edited: Sep 27, 2011
2. Sep 27, 2011

### PeterO

Re: Projectile motion question

(a) looks good.

(b) is not appropriate. You know how high it went, you can calculate how long - in time - that took. It has to come down again, so there is some more time.
All that while the cart had been travelling at its constant speed, so will covered a distance determined by its speed and the time.
(c) logical method - if your arithmetic is correct tat should be right.
(d) & (e) both good.

3. Sep 27, 2011

Re: Projectile motion question--nothing special.

Thanks Peter! I will work on part b tomorrow, I have to sleep now.

4. Sep 27, 2011

Re: Projectile motion question--nothing special.

I guess I can't do arithmetic!

x sin x = 43.1
x cos x = 38.0

(43.1)^2+(38)^2 =3301.61

sq.root = 57.45963801

43.1/57.4596 = .75009

sin^1(.75009)= 48.598

angle = 48.6 degrees above the horizontal. but wait! its not. I messed up somewhere. Thanks.

Maybe the "above the horizontal" doesn't mean this angle?

Last edited: Sep 27, 2011
5. Sep 27, 2011

### PeterO

Re: Projectile motion question--nothing special.

There is nothing wrong with your original answer to (c) - I just checked it. I couldn't be bothered checking last time as it looked like you had no difficulty. The only possibility was a slip in arithmetic, which there was none.

6. Sep 27, 2011

Re: Projectile motion question--nothing special.

I typed my answer for c into the homework, and it said I was wrong. I received 334.2 m for part B, but was wrong again.

I have two attempts remaining for each now, but I have no idea what went wrong.

Does the observer at rest have anything to do with it?

7. Sep 27, 2011

Re: Projectile motion question--nothing special.

OMG- so stupid. The Vox is 33.0, not 38.0.

Now I'm depressed..

Thanks Peter.

My new angle is 52.67

My new Vo is 54.2033

and thats right.

Last edited: Sep 27, 2011