# Angle at which canon can be fired

• gwizz
In summary: The angle is 7.8.In summary, the homework equation for the angle above the horizontal that a canon should be fired at is v=vy+ayt, where v is the vertical initial velocity and y is the height above the horizontal. The equation simplifies to v=vy+g, where g is the acceleration due to gravity.
gwizz

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

Not sure how to start this.

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:
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.

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)

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

gwizz said:
SammyS, if i use the Y equivalent to the equation i had, does that mean what i did was incorrect?
Probably it's wrong.

How did you get it from the information you have?

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?

gwizz said:
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?

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?

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?

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.

$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
$\alpha = 22°$
$\alpha = 89°$

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.

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

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

## What is the angle at which a canon can be fired?

The angle at which a canon can be fired depends on various factors such as the weight and size of the projectile, the distance to the target, and the desired trajectory. Generally, the angle can range from 0 degrees (horizontal) to 90 degrees (vertical).

## What is the optimal angle for firing a canon?

The optimal angle for firing a canon depends on the specific situation and goals. For maximum distance, the optimal angle is typically between 45 and 60 degrees. For precision, the angle may vary depending on the target's location and obstacles in the trajectory path.

## Can a canon be fired at any angle?

Technically, a canon can be fired at any angle. However, there are limitations based on the canon's design and capabilities. For example, some canons may have a limited range of motion and can only be fired at certain angles.

## What happens if a canon is fired at a too high or too low angle?

If a canon is fired at a too high or too low angle, the projectile may not reach its intended target or may not have enough force upon impact. This can result in a failed attempt or cause collateral damage.

## How does the angle of a canon affect the trajectory of the projectile?

The angle of a canon plays a crucial role in determining the trajectory of the projectile. A higher angle can result in a steeper trajectory and vice versa. The angle also affects the distance the projectile can travel and the amount of force upon impact.

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