How to derive the range equation starting with the equations of 1-d motion?

Question

How to derive the range equation starting with the equations of 1-d motion?
@=the angle
o=initial
v(xo)=initial velocity in the horizontal direction and v(yo)= initial velocity int he vertical direction
range equation to derive: R=(vo^2 * sin2@o)/g
1-motion equations- x= xo + v(xo)*t and y= yo + v(yo)*t-(1/2)*g*t^2
other 1-d motion equations: v(xo)=vo(cos@o) and v(yo)=vo(sin@o) and 2sin@cos@=sin2@.

I keep trying but iIm just stumped. Its confusing. They never taught us this in our physics class, and i cant find it in our book. Anyone have an idea?

Answer 1

What you appear to be asking is, how to derive the range of a projectile across level ground in a uniform gravitational field, g, launched from ground level at a given angle and initial speed, neglecting air resistance.



What you have to do is derive the equation of the trajectory (y vs x!), which will be a parabola, and find the non-zero root, because what you want to know is x, when the projectile is on the ground, i.e., when y=0. This happens twice -- at launch and at impact. What you have are equations of motion for the x- and y-components of the trajectory, in terms of the elapsed time, t. This gives you the trajectory in parametric form, the parameter being t.



Since the x-equation is linear, you can solve it for t in terms of x, plug that result into the y-vs-t equation, and get y in terms of x. This will be a quadratic equation, one of the roots of which is x=0 (the launch point!). Because you'll have a quadratic with one root=0, the other root will be especially transparent, and will be the answer you seek.

Answer 2

Terrific a question that tests your knowledge of the physics and not your ability to memorize equations. Love it, kudos to your instructor. Anyway...



We start with R = Ux T; where R is the range (as in range equation), Ux = U cos(theta) is the horizontal launch speed, where U is the launch speed and theta is the launch angle re the horizontal, and T is total time in the air. We need to find T in terms of U and theta.



T = tu + td; where tu = Uy/g is the time to rise from launch height to max height in the trajectory. Uy = U sin(theta) is the vertical launch speed.



Lacking any forces other than gravity and assuming the launch and impact heights are the same heights, we can argue that tu = td = t because it takes just as long to fall to Earth from max height as it does to reach max height from launch. In which case we have T = tu + td = t + t = 2t = 2 Uy/g = 2 U sin(theta)/g.



We now have T in terms of U and theta.



So R = Ux T = U cos(thetea) 2 U sin(theta)/g = (U^2/g) 2 cos(theta) sin(theta) = (U^2/g) sin(2theta). QED



Maybe "they" didn't teach you the derivation, but I'm betting they taught you the physics, which you should know if you were paying attention. If you know the physics, you can always derive the equations.

Answer 3

it is the commencing equation for deriving the 2nd equation of action: s= ut + a million/2 at^2. permit's start up with the definition of distance: distance travelled = universal velocity x time. We define universal velocity as: v(av) = (v+u)/2 according to this being the arithmetic thank you to calculate averages or mean. putting this in our definition of distance: s = v(av)/2 s= (v+u)t/2 desire this facilitates! further: to apply this to derive the 2d equation of action you proceed as follows: positioned v=u+at on your derived effect: s= (u+at+u)t/2 Simplifying and commencing the bracket, you get: s= ut + (at^2)/2