# Homework Help: Simple 2D ballistics

1. Dec 4, 2009

### Torquescrew

Well, finals are coming around. I'm reviewing stuff I thought I learned at the beginning of the semester only to find out that I don't remember what I'm doing. It's pretty easy until I forget everything.
I think I have a handle on this, but I just want some verification to make sure I'm doing it right.

1. The problem statement, all variables and given/known data
A man stands on the roof of a building that is 30.0 m tall and throws a rock with a velocity of magnitude 60.0 m/s at an angle of 33.0° above the horizontal. Calculate a) the maximum height above the roof reached by the rock; b) the magnitude of the velocity of the rock just before it strikes the ground; c) the horizontal distance from the base of the building to the point where the rock strikes the ground.

2. Relevant equations
I'm a big fan of using these:
Vf=Vo+a(t)
(delta)x=Vo(t)+.5(a)(t^2)
(delta)x=[Vf^2 - Vo^2]/2a

and some basic trig

3. The attempt at a solution

For (a), I chopped up my initial vector into x and y components.
Vox = 50.3202 m/s
Voy = 32.6783 m/s

Then, I figured I'd not only see how far up it would go, but how long it would take to get there.
For time, I used:
0=32.6783{m/s}-9.81{m/s^s}*t
I wound up with 3.3113 seconds

Next, I got the same result of 54.4278 meters (above the roof) whether I used this equation
(delta)x=Vo(t)+.5(a)(t^2) <--using the time I had previously solved for
or this one
(delta)x=[Vf^2 - Vo^2]/2a <--ignoring time altogether

Then, for (b) I added the 54.4 meters to my initial 30 and solved for a new time (this one for it's downward movement) with this one again:
(delta)x=Vo(t)+.5(a)(t^2)
84.4...=0+(.5*9.81*t^2)
Netted myself another 4.14881 seconds

Using this new time, I solved for my final y velocity with
Vf=Vo+a(t)
Vf=0+9.81*4.14881
I wound up with 40.6998 m/s
I verified by using this formula:
Vf^2=Vo^2 + 2a(delta x)
or rather
Vf^2=0+(2*9.81*84.4...)

Then, I used the handy formula for building one vector out of 2:
[(x^2)+(y^2)]^(1/2) <-- somebody is gonna' have to teach me how to type a "square root" symbol
[(50.3202^2 m/s) + (40.6998^2 m/s)]^(1/2)
I now have a vector magnitude of about 64.7 m/s

Lastly, for (c), I added my two times together
4.14881+3.3113
Very nearly 7.5 seconds in flight.

I multiplied my initial x vector by the flight time to see how far it flew before it landed.
50.3202 m/s * 7.47993 s

So it looks like this guy threw a rock 376.4 meters.

That's over three and a half football fields.
Either this guy has one heck of an arm (60 meters a second initial velocity? Madness!), or I royally jacked this up.

Am I crazy? More importantly, are my calculations correct?
Something just seems out of place and I can't put my finger on it.

2. Dec 4, 2009