We can use the range equation to calculate how far the arrow travels until it gets back its initial height. However, we then need to calculate how far the arrow travels from there until it hits the ground. So, beginning with the range equation:

Plugging in our values:

Now we need to calculate the horizontal distance that the arrow travels after this point. First, let's calculate how long it takes to hit the ground. At this point, the arrow has the same speed that it began with, but its velocity is now directed at  below the horizontal. Therefore, its vertical component:

Then using the following kinematic equation:

Plugging in our values:

Rearranging:

Then using the quadratic formula, we get:

 or 

Obviously, we can't have a negative time, so the first option is the correct one. We can then multiply this by the horizontal component of the arrow's velocity, which is constant since we are neglecting air resistance, to get the rest of the horizontal distance traveled:

Then adding our two distances together to get our total distance traveled, we get:

The key component of the problem statement is that the arrow is shot from a height of 1m and hits a target that is also at 1m. Therefore, since we are neglecting air resistance, we can conclude that the arrow hits the target at the same speed that it begins with, except that the velocity is now oriented  below the horizontal. Thus, the arrow hits the target with the same kinetic energy that it begins with, and the target absorbs all of this energy. Therefore, we can say:

Note that with the information given in the problem statement, the angle that the arrow was shot from and the distance the target was from where the arrow was shot are both insignificant.

Since the problem statement tells us that the bow can be modeled as a spring, we can use the following expression to determined how much energy is stored in the bow:

Where  is the spring constant, and  is the distance the string of the bow is pulled from equillibrium. Plugging in our values, we get:

Then we can assume that all of this energy is transferred to the arrow as kinetic energy:

Rearranging for initial velocity:

Plugging in our values:

Then we can use the range equation to calculate the range of the arrow. Since we were given no initial or final heights in the problem statement, we will assume that the initial and final heights are the same; thus, the range equation is applicable:

We can use the following kinematic equation to determine the arrow's initial vertical velocity:

Rearranging for initial velocity, and eliminating final velocity:

Substituting in values:

Then using the following equation to determine total initial velocity:

Rearranging for initial velocity:

Plugging in our values:

We can use the range equation to solve this problem:

Now let's start to rearrange for theta:

Let's stop here and develop an expression for initial velocity in terms that we know"

Rearranging:

Now let's plug this back into the range equation:

Then we can use the following trig identity:

Plugging this into our expression:

Now we can cancel out a cos function to get:

And then:

Rearranging for theta:

We have values for each of these variables, so time to plug and chug:

In order for the object not to gain velocity, all the forces acting on it need to cancel out.  The normal force  will already cancel with the perpendicular force due to gravity () by definition.  In order to find the correct coefficient of friction, we set the magnitude of the force due to friction  equal to that of the parallel force due to gravity :

The mass and gravity cancel here.

Using

Finding the angle at which the car launches at:

Determining y-component of velocity:

Using conservation of energy in a closed system:

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

Solving for

Plugging in values:

Finding the angle at which the car launches at:

Determining component of velocity:

Using conservation of energy in a closed system:

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

Solving for

Plugging in values:

Determine the velocity in the y direction when the car hits the ground

Solving for

Plugging in values:

Now, breaking the jump into two parts, the "up" and the "down" and using

Up:

Down:

Finding the angle at which the car launches at:

Determining component of velocity:

Using conservation of energy in a closed system:

Initially, the car has both kinetic and gravitational potential energy in the y direction, but after, it only has gravitational potential.

Solving for

Plugging in values:

Determine the velocity in the y direction when the car hits the ground

Solving for

Plugging in values:

Now, breaking the jump into two parts, the "up" and the "down" and using

Up:

Down:

Multiplying by the horizontal velocity: