How do you derive the correct projectile motion formula?

[gmy5011]

Homework Statement

A jaguar(A) leaps from the origin at a speed of v₀ = 6 m/s and an angle β = 35° relative to the incline to try and intercept the panther(B) at point C. Determine the distance R that the jaguar jumps from the origin to point C. given the the angle of the incline is θ = 25°.

Homework Equations

a = dv/dt

The Attempt at a Solution

I know how to solve this problem by just looking up the constant acceleration formula and translating the velocity and R into cosines and sines. My question is, How do we know that we have to derive the equation s = s_0 +v_0(t) +.5a(t^2)? When I first tried it, and since it was trying to relate velocity and distance I thought I would use the derivation a = (dv/dx)(dx/dt) = v(dv/dx). Once I integrated it out I got a(x - x_0) = (v^2)/2 - (v_0^2)/2... why doesn't this work for solving the problem? How do you know from the beginning that you're going to have to use the general form I showed above when we are not given any info about time in the problem statement? Let me know if I need to clarify my question at all.

 

[mfb]

How do we know that we have to derive the equation s = s_0 +v_0(t) +.5a(t^2)?

You do not have to, but it is useful.

Once I integrated it out I got a(x - x_0) = (v^2)/2 - (v_0^2)/2... why doesn't this work for solving the problem?

How does the height-dependence of the velocity help?

How do you know from the beginning that you're going to have to use the general form I showed above when we are not given any info about time in the problem statement?

Experience. In doubt, use the general formula.

 

[gmy5011]

Thanks for the reply! Do we always have to use the general form for projectile motion since it is 2-D. I now kind of think we do, because we are going to need a way to relate the equations in both directions, so we are going to have to use a form with time because that's the variable that is the same in both directions, is that correct? Does that make sense?

 

[mfb]

There are always multiple methods to solve the problem, but that one is the easiest.