Angle at which canon can be fired

[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.

 

[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! )

 

[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 + volt + (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.

 

[SammyS]

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)

 

[gwizz]

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

 

[SammyS]

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?

 

[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?

 

[SammyS]

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:

(V_F)_Y = (V₀)_Y + a_Y · t

Y_F = Y₀ + (V₀)_Y · t + (1/2)a_Y · t²

What is a_Y ?​

For the X-components:

(V_F)_X = (V₀)_X + a_X · t

X_F = X₀ + (V₀)_X · t + (1/2)a_X · t²

But gravity acts only vertically, so what is a_X ?
That will simplify the equations for the X-components.​

What do the equations become when you replace a_X & a_Y with the appropriate values for acceleration?

 

[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?

 

[SammyS]

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 X_F to find t in terms of θ.

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

 

[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
$\alpha = 22°$
$\alpha = 89°$

 

[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.

 

[jaumzaum]

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

 

[gwizz]

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