Need some help on a spring cannon problem

[TexasBB12]

Homework Statement

Launch a .8 kg ball from a spring cannon in order to hit a target 25m away, the cannon fires at 45 degrees relative to the ground, what is the initial velocity necessary for the ball to make the distance. spring constant is 425 N/m.

Homework Equations

fspring=-kx

The Attempt at a Solution

I really don't know how to get started on this.

 

[rock.freak667]

Start by finding the velocity. Assume the velocity is 'v' at an angle of 45°. How would you find the range (or how far the mass travels) of the ball? What relevant equations will you need?

 

[TexasBB12]

vx=vcos45 vy=vsin45
distance is 25m
The problem is set up so i can't use energy to solve for the velocity
the only equation i can really think of is vf2=vi2+2ax
would i solve for the y component using a=-9.8??

 

[rock.freak667]

vx=vcos45 vy=vsin45
distance is 25m
The problem is set up so i can't use energy to solve for the velocity
the only equation i can really think of is vf2=vi2+2ax
would i solve for the y component using a=-9.8??

If you consider vertical motion, when the ball hits the ground again, the displacement will be zero. So you can get the time of flight from that in terms of v.

Horizontally it travels 25m. So 25 = ?

 

[TexasBB12]

25m=0+vi(t) +1/2axt2
so i have t=25/vcostheta, this is where i don't know how to solve for time

 

[rock.freak667]

25m=0+vi(t) +1/2axt2
so i have t=25/vcostheta, this is where i don't know how to solve for time

You are confusing the vertical motion with horizontal motion.

Vertically when the ball reaches the 25m mark, the overall vertical displacement is zero.

So if you use s=v_i-1/2gt², what is t in terms of v_i?


Horizontally, the horizontal distance is nothing but initial horizontal velocity x time.

Your horizontal distance is your range, 25m. So now you should be able to solve for v_i

 

[TexasBB12]

vertically t=vysintheta/4.9 so 25=vi- vysintheta/4.9??
I'm just really needing to figure this out as I'm on a time crunch

 

[rock.freak667]

vertically t=vysintheta/4.9 so 25=vi- vysintheta/4.9??
I'm just really needing to figure this out as I'm on a time crunch

Ok you got vertically that t = v_isinθ/4.9. That gives you the time for the entire flight.


Now horizontally the distance the mass travels is 25m. Horizontal distance = initial horizontal velocity x t


What is your initial horizontal velocity?

 

[TexasBB12]

vi=13.16??

 

[rock.freak667]

vx=vcos45 vy=vsin45
distance is 25m
The problem is set up so i can't use energy to solve for the velocity
the only equation i can really think of is vf2=vi2+2ax
would i solve for the y component using a=-9.8??

vi=13.16??

See the part in bold, those are your initial horizontal and vertical velocities respectively.

 

[TexasBB12]

25=.707vi x (.707vi/4.9) ??

 

[rock.freak667]

25=.707vi x (.707vi/4.9) ??

Right yes, but remember that your '0.707' is really 1/√2, for a more accurate answer.

So you can now find v_i.


To get the how much the spring is compressed, think of when the spring is compressed, it has elastic potential energy and as it is released, all of that is converted entirely into kinetic energy.