# Projectile range

1. Dec 31, 2013

### negation

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

After a short engine firiing, an atmosphere-probing rocket reaches 4.6km/s. If the rocket must land within 50km of its launc site, what's the max allowable deviation from a vertical trajectory.

2. Relevant equations
none

3. The attempt at a solution

Given by the book:

x = (vi^2/g)sin 2Θ

50,000m = (vi^2/-9.8)sin 2Θ

sin2Θ = (gx/vi^2) = 0.0232

Again this is really frustrating, I take g = -9.8 and x = 50,000m, vi = 4600m/s
but I get -0.0232.

What is wrong here?
If I use a positive g, I get 0.0232 but that would be rubbish.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 31, 2013

### Curious3141

Why would it be "rubbish"?

I suggest you check the derivation of the range formula. That should tell you why $g$ should be taken as a positive number (just the magnitude) here.

3. Dec 31, 2013

### negation

I understand how to derive the equation for the range formula. What I don't understand is that the g in the equation was not specified as a magnitude so why should I be using the magnitude and not the vector g?

4. Jan 1, 2014

### Curious3141

So, in the derivation, what is the exact expression for the vertical position at time $t$?

5. Jan 2, 2014

### negation

I have understood why g = 9.8 instead of -9.8ms^-2

6. Jan 2, 2014

### Curious3141

Good. It is because the equation for the vertical displacement explicitly includes a negative sign before $g$, i.e. $\displaystyle y = v\sin\theta{t} -\frac{1}{2}gt^2$. Hence $g$ here refers to the magnitude only.