# Homework Help: An artillery shell is fired I don't know!

1. Oct 18, 2007

### Cowtipper

1. The problem statement, all variables and given/known data
An artillery shell is fired with an initial velocity of 300 m/s at 55.0 degrees above the horizontal. It explodes on a mountainside 42.0 seconds after firing. What are the x and y coordinates of the shell where it explodes, relative to it's firing point?

We know that it is fired at 300 m/s. It is launched at a 55 degree angle and it hits after 42 seconds of flight.

2. Relevant equations

See below.

3. The attempt at a solution
--x component--

velocity: 172.1 m/s
distance: 7228.2 meters
time: 42 seconds
acceleration: 0 m/s/s

velocity = 300cos55
distance = 172.1 m/s x 42 seconds
time = given
acceleration = always zero

--y component--

velocity: 245.7 m/s
distance: ? - This is what I'm trying to figure out.
time: 25.0 seconds
acceleration: -9.81 m/s/s

velocity = 300sin55
distance = ??
time = 245.7 meters / 9.81 m/s/s
acceleration = -9.81 m/s/s

I am confused. Thank you very much.

2. Oct 19, 2007

### learningphysics

d = v1*t + (1/2)at^2

apply this to the vertical direction.

3. Oct 19, 2007

### Coto

"velocity: 245.7 m/s
distance: ? - This is what I'm trying to figure out.
time: 25.0 seconds
acceleration: -9.81 m/s/s

velocity = 300sin55
distance = ??
time = 245.7 meters / 9.81 m/s/s
acceleration = -9.81 m/s/s"

You show your time to be 25.0 seconds, but this isn't quite correct. Essentially, you've calculated the amount of time that it would take for the rocket to reach its maximum height (i.e. vertical velocity = 0). However your total time in the air is still 42 seconds. This means your rocket now continues downward, and due to your coordinate system, it's portrayed as a negative velocity. Thus, you can now use the formula given in the last post, with your a=-9.81 , t = 42s, and v(initial) = 245.7m/s. You should pop out with something around 1.6km, a reasonable answer considering the average (Rocky) mountain is around 2.5km.

4. Oct 19, 2007

Thanks!