# Angle at which canon can be fired

1. Jul 7, 2011

### gwizz

1. The problem statement, all variables and given/known data
canon is firing at a mountain and it want the angle above the horizontal which it should be fired

muzzle speed= 1000m/s
x=2000m
y=800m

2. Relevant equations

vxi=vicos(theta) vyi=visin(theta)

3. The attempt at a solution

Not sure how to start this.

2. Jul 7, 2011

### tiny-tim

welcome to pf!

hi gwizz! welcome to pf!

(have a theta: θ and try using the X2 icon just above the Reply box )

write out the https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations for the x and y directions (separately, with a = 0 and -g) …

that gives you two equations with two unknowns (v and t), so eliminate t to find v …

what do you get?

(btw, a canon is a priest … you mean cannon! )

Last edited by a moderator: Apr 26, 2017
3. Jul 7, 2011

### gwizz

sorry, cannon. also, pardon for the equations, idk how to use the wrap stuff with the tool.

so for X i used the kinematic equation: Xf=Xo + Vot + (1/2)at^2 and got t=2 seconds then plugged that in the Y kinematic equation Vy=Voy + ayt and got a vertical initial velocity of Voy= 19.6 m/s. Is this correct so far?
If so, now what? thanks.

4. Jul 7, 2011

### SammyS

Staff Emeritus
Acceleration in x direction is zero, so the x component of velocity is constant.

Acceleration in y direction is -g = -9.81 m/s2, or whatever approximation is acceptable for you class.

y = y0 + (v0)yt - (1/2)g(t2)

5. Jul 8, 2011

### gwizz

SammyS, if i use the Y equivalent to the equation i had, does that mean what i did was incorrect?

6. Jul 8, 2011

### SammyS

Staff Emeritus
Probably it's wrong.

How did you get it from the information you have?

7. Jul 8, 2011

### gwizz

i just thought those equations were the one to use based off tiny-tim's help. What do i do exactly, with what equations to get started then?

8. Jul 8, 2011

### SammyS

Staff Emeritus
Use those equations along with what you know about the acceleration due to gravity.

For the Y-components:
(VF)Y = (V0)Y + aY · t

YF = Y0 + (V0)Y · t + (1/2)aY · t2

What is aY ?​

For the X-components:
(VF)X = (V0)X + aX · t

XF = X0 + (V0)X · t + (1/2)aX · t2

But gravity acts only vertically, so what is aX ?
That will simplify the equations for the X-components.​
What do the equations become when you replace aX & aY with the appropriate values for acceleration?

9. Jul 8, 2011

### gwizz

ay is -9.8m/s^2 due to gravity and ax is 0 because it is always zero in the x direction. is my velocity correct for Vyin my previous post?

10. Jul 8, 2011

### SammyS

Staff Emeritus
I don't think you can find (V0)Y unless you know the launch angle. (That is, unless you're a savant.)

You do know V0, because it's given (muzzle speed). Use trig to relate the launch angle, θ, and V0 to (V0)X & (V0)Y.

Use the kinematic equation for XF to find t in terms of θ.

Plug that result into the kinematic equation for YF. Solve for θ. It's kind of tricky.

11. Jul 8, 2011

### jaumzaum

$Vx = V cos\alpha$

$Voy = V sen\alpha$

$x = Vx .t = V cos \alpha.t$

$y = Voyt - gt² /2 = V sen\alpha t - gt²/2$

Isoling t

$t = \frac {x} {V cos\alpha}$

$y = x tan\alpha - g x²/2V²cos² \alpha$

Replacing we find
$\alpha = 22°$
$\alpha = 89°$

12. Jul 9, 2011

### gwizz

ya it is tricky, thanks for the help. and Jaumzaum, i got an angle of like 21 but the answer is 7.8 or something like that.

13. Jul 10, 2011

### jaumzaum

First of AL, if g is down, tan > 800/1000, angle > 21º, otherwise muzzle will pass beneath

14. Jul 11, 2011

### gwizz

well i'm just restating the answer i got from the answer sheet