# Projectile Motion Cannon Angle

1. Sep 20, 2008

### specwarop

Okay another projectile motion question. I'm a bit confused by how to work this out.

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A cannon will a muzzle speed of 1000m/s is used to start an avalanche on a mountain slope. The target is 2000m from the cannon horizontally and 800m above the cannon. At what angle, above the horizontal, should the cannon be fired?
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I was thinking that i would have to work out the angle to which the cannon should fire if the target was at its vertical level, 2000m away horizontally. Then somehow (i don't know how), using this angle in a further equation, that i don't know of, to include the difference in height?
Can anyone help?
Regards.

2. Sep 20, 2008

### Kurdt

Staff Emeritus
The trick (as always) is to consider each component of motion separately. you should then have two equations with two unknowns (time and angle). You can substitute out on of the unknowns and find the other.

3. Sep 20, 2008

### prateek_34gem

You should always solve the projectile problem like this :

Let it be projected at angle 'x' with horizontal. Then at instantaneous time 't' :

Along X- Direction

u=1000cosx
a=0
v=1000cosx
s=1000cosxt

Along Y- Direction

u=1000sinx
a= -g
v=1000sinx-gt
s=1000sinxt-1/2gt2

Your displacement along x-direction will be
1000cosxt
According to question
1000cosxt=2000----------(1)

Your displacement along y-axis will be
1000sinxt-1/2gt2
According to question,
1000sinxt-1/2gt2=800------(2)

Now you can solve as you have two equations( 1 & 2) and two variables x and t.