Firing cannons projectile motion

[Psychros]

Homework Statement

A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)

Homework Equations

Time= 0=v₀T-½gT² T<0
t= (v_Oy+√(v_Oy²-2gy))/g
distance/range x=x₀+V_0xt
x/y velocity- v_0x=Vcos(θ); v_0y=vsin(θ)

The Attempt at a Solution

This is a practice problem (with different values than original), and the provided answer is 32.9m/s

i set the origin at the cannon
Since the velocity magnitude = √(Vcos(θ)²+Vsin(θ)²)

I tried variations along the line of V₀cos(30)²+V₀sin(30)²= 1.3 V₀²
in order to split the velocity into xy components.

Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.

Any and all help very appreciated, thank you.

 

[SteamKing]

The Attempt at a Solution

This is a practice problem (with different values than original), and the provided answer is 32.9m/s

i set the origin at the cannon
Since the velocity magnitude = √(Vcos(θ)²+Vsin(θ)²)

I tried variations along the line of V₀cos(30)²+V₀sin(30)²= 1.3 V₀²
in order to split the velocity into xy components.

Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.

Any and all help very appreciated, thank you.

I'm not sure what you are trying to do here. The angle at which the projectile strikes the ground may not be equal to the initial firing angle, i.e. 30 degrees.

The best way to approach this problem is to find the maximum altitude which the projectile reaches after it is fired. At this point, the vertical component of the velocity is zero. After reaching the maximum altitude, the projectile falls under the force of gravity. Knowing the magnitude of the final velocity, you should be able to determine the duration of the fall and the resulting vertical velocity component. Remember, since there is no air resistance, the horizontal component of the velocity remains unchanged during the entire flight of the projectile.

 

[Psychros]

I'm not sure what you are trying to do here. The angle at which the projectile strikes the ground may not be equal to the initial firing angle, i.e. 30 degrees.

The best way to approach this problem is to find the maximum altitude which the projectile reaches after it is fired. At this point, the vertical component of the velocity is zero. After reaching the maximum altitude, the projectile falls under the force of gravity. Knowing the magnitude of the final velocity, you should be able to determine the duration of the fall and the resulting vertical velocity component. Remember, since there is no air resistance, the horizontal component of the velocity remains unchanged during the entire flight of the projectile.

That is brilliant, thank you. However, Without knowing distance (x), time, or hard value for velocity (since it's expressed as a relation to the initial velocity) I'm not sure how to go about this.

 

[SammyS]

Homework Statement

A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)

Homework Equations

Time= 0=v₀T-½gT² T<0
t= (v_Oy+√(v_Oy²-2gy))/g
distance/range x=x₀+V_0xt
x/y velocity- v_0x=Vcos(θ); v_0y=vsin(θ)
...

Try using conservation of energy.

 

[Psychros]

Try using conservation of energy.

Hmmm I don't think we've learned that yet..

 

[SammyS]

Hmmm I don't think we've learned that yet..

In that case,

What can you conclude about the x component of the velocity?


What kinematic equations do you know regarding the vertical component of the velocity ?

 

[Psychros]

X component will be the same through it's entirety.

y=y₀+v_0y(t)+½at²
or
V_y=V_0y+gt
V_y=1.3V_0y-9.81t
But Theoretically if we were starting at the maximum, V_0y would = 0,

V_y/-9.81=t?

I'm getting stuck down either avenue, I do think I'm missing something intuitive here..

 

[SammyS]

X component will be the same through it's entirety.

y=y₀+v_0y(t)+½at²
or
V_y=V_0y+gt
1.3V_y=V_0y-9.81t
1.3V_y=0-9.8t
V_y=(-9.8t)/1.3

I'm getting stuck down either avenue, I do think I'm missing something intuitive here..

v_y ≠ 1.3 v_0y . I'm assuming that you mean v_y to be the y component of the final velocity.

How is speed obtained from components of velocity ?


Also, there's another kinematic equation. One that doesn't involve time, t.

Edited:
[STRIKE](v_y)² = (v_0y)² - g(Δy)[/STRIKE]

(v_y)² = (v_0y)² - 2g(Δy)

 

[Psychros]

Oh yes, i made a false assumption about the y component.
Do you mean speed as in magnitude of velocity?

Also, should (v_y)² = (v_0y)² - g(Δy)
have a -2g(Δy)?

I apologize for being slow, but I'm not sure how to use this either since I don't know the change in height from the maximum

 

[SammyS]

Oh yes, i made a false assumption about the y component.
Do you mean speed as in magnitude of velocity?

Also, should (v_y)² = (v_0y)² - g(Δy)
have a -2g(Δy)?

I apologize for being slow, but I'm not sure how to use this either since I don't know the change in height from the maximum

Right. There should be a 2 in there.


Then, ...

I'll ask again,

How is speed obtained from components of velocity ?​

 

[Psychros]

Since speed is the magnitude of velocity,
by components of velocity I imagine you mean V_x and V_y,
or √(Vcos(θ)²+Vsin(θ)²)?

 

[SammyS]

Since speed is the magnitude of velocity,
by components of velocity I imagine you mean V_x and V_y,
or √(Vcos(θ)²+Vsin(θ)²)?

Yes. Of course, sin²(θ) + cos²(θ) = 1

Anyway, $\ v_0=\sqrt{(v_{0x})^2+(v_{0y})^2}\ ,\$ so that $\ (v_0)^2=(v_{0x})^2+(v_{0y})^2\ .\$

And in general $\ v^2=(v_{x})^2+(v_{y})^2\ .\$ Right?


Now, what do you know regarding the x component of velocity for this projectile?

 

[Psychros]

sorry, that's what i meant when i put √(V⁰cos(θ)²+V₀sin(θ)²), that it's = V₀.

I know that it doesn't change, but I'm not sure how to find its' magnitude in this situation.

 

[SammyS]

sorry, that's what i meant when i put √(V⁰cos(θ)²+V₀sin(θ)²), that it's = V₀.

I know that it doesn't change, but I'm not sure how to find its' magnitude in this situation.

So then $v_{0x}=v_x\$ .

Regarding your kinematic equation with the $(v_{y})^2\,$, what is it missing for it to have v² and v₀² in it ?