Deriving an equation for a projectile launched on a ramp using two springs

Question

I had a physics lab where I had to build a sling shot sort of devices on a ramp. Basically, I have a ramp, with two springs attached on the end and the two strings are attached to a piece of wood. I put something in front of the piece of wood, pull the wood back, and let go. This launches the projectile.

What I am having trouble with is deriving equations. Here's what I mean:

This is directly from the assignment:

Derivation: Derive a set of equations to use to determine how far back (x) you need to pull a mass, m, in your slingshot to launch it a distance R. Your derivations will also include the spring constant, k, acceleration due to gravity, g, the launch angle, θ, and the height at which the projectile is launched, H. You should obtain 2 equations: One that gives the launch speed, v, in terms of R, θ, and H, and the other that allows you to solve for x in terms of k, m, g, θ, and v. Leave values in your equations as variables. On launch day you will plug in the values as needed. You will know k from your lab. R, and m you will be given on launch day. You can choose any value for θ that you desire. You will be able to calculate a value for v from the first equation to use in the second one.

The problem is, I don't know how to derive the equations. Here's some information I have from testing:

Launching a 270 g object takes 1.76 sec to hit the ground. Also, the distance it traveled horizonatally was 1.4 m. Other information for the ramp is in the diagram. K value and such is unknown. X value when launched was 10 cm for above information.

Answer 1

As this is a HW&W type question I'll only try and point to a few helpful principles.

If you choose a suitable reference frame, above the blue $x,y$ coordinate system, things tend to become clearer.

As the derivation becomes otherwise very difficult, we'll assume absence of friction, drag forces and such.

Now bear in mind that motions in the $x$ and $y$ directions are completely independent from each other and you can take great advantage of that, as follows.

If the object leaves the ramp at $v_0$, then decompose that velocity vector into an $x$ and a $y$ component:

$v_{0,x}=v_0\cos \theta$

$v_{0,y}=v_0\sin \theta$

Due to gravity, the $y$ component will now evolve as:

$$v_y(t)=v_{0,y}-gt$$

And the $y$-position is given by:

$$y=v_{0,y}t-\frac12 gt^2$$

Using the landing coordinate $y=-0.49\:\mathrm{m}$ and the time in the air, $t=1.76\:\mathrm{s}$, you can calculate $v_{0,y}$ and thus $v_0$.

To calculate the $k$ values of the springs, use the work-energy theorem. When the springs are pulled back, they contain potential energy, equal to half the product of $k$ and the spring extension squared. When released (in the absence of friction) that energy is converted to kinetic energy of the object $\frac12 mv_0^2$. So the spring constant can be calculated from $v_0$.

Regarding the distance travelled in the $x$-direction, it's simply given by: $x=v_{o,x}t$ with $t$ the flight time, because gravity doesn't act in the $x$ direction.

I hope this helps.