Explain why mass does not enter into the range equation? Solution Ideal projecti

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Ideal projectile motion Ideal projectile motion assumes that there is no air resistance. This assumption simplifies the mathematics greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are small. Ideal projectile motion is also a good introduction to the topic before adding the complications of air resistance. Derivations 45 degrees goes the furthest. Flat ground Range of a projectile (in vacuum). First we examine the case where (y0) is zero. The horizontal position of the projectile is x(t) = v t cos heta In the vertical direction y(t) = v t sin heta - rac{1} {2} g t^2 We are interested in the time when the projectile returns to the same height it originated at, thus y = v t sin heta - rac{1} {2} g t^2 By factoring: t = 0 or t = rac{2 v sin heta} {g} The first solution corresponds to when the projectile is first launched. The second solution is the useful one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation yields x = rac {2 v^2 cos heta , sin heta } {g} Applying the trigonometric identity sin(x+y) = sin x , cos y + sin y , cos x If x and y are same, sin 2x = 2 sin x , cos x allows us to simplify the solution to d = rac {v^2 sin 2 heta}{g} Note that when (?) is 45°, the solution becomes d = rac {v^2} {g} Uneven ground Now we will allow (y0) to be nonzero. Our equations of motion are now x(t) = v t cos heta and y(t) = y_0 + v t sin heta - rac{1}{2} g t^2 Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with) 0 = y_0 + v t sin heta - rac{1} {2} g t^2 Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation t = rac {v sin heta} {g} pm rac {sqrt{v^2 sin^2 heta + 2 g y_0}} {g} The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is t = rac {v sin heta} {g} + rac {sqrt{v^2 sin^2 heta + 2 g y_0}} {g} Solving for the range once again d = rac {v cos heta} {g} left [ v sin heta + sqrt{v^2 sin^2 heta + 2 g y_0} ight] Maximum range Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows). For cases where the projectile lands at the same height from which it is launched, the maximum range is obtained by using a launch angle of 45 degrees. A projectile that is launched with an elevation of 0 degrees will strike the ground immediately (range = 0), though it may then bounce or roll. A projectile that is fired with an elevation of 90 degrees (i.e. straight up) will travel straight up, then straight down, and strike the ground at the point from which it is launched, again yielding a range of 0. The elevation angle which will provide the maximum range when launching the projectile from a non-zero initial height can be computed by finding the derivative of the range with respect to the elevation angle and setting the derivative to zero to find the extremum: rac { dR } { d heta} = rac {v^2} {g} left[ cos heta left( cos heta + rac {sin heta , cos heta} {sqrt {sin^2 heta + C} } ight) - sin heta left( sin heta + sqrt { sin^2 heta + C } ight) ight] where C = rac {2 g y_0} { v^2} and ,!R = horizontal range. Setting the derivative to zero provides the equation: cos^2 heta + rac {sin heta , cos^2 heta} {sqrt {sin^2 heta + C} } - sin^2 heta - sin heta sqrt {sin^2 heta + C} = 0 Substituting u = cos^2 heta and 1-u= sin^2 heta produces: u + rac {u sqrt {1-u}} {sqrt {1-u+C}} - (1-u) - (sqrt {1-u}) sqrt {1-u+C} = 0 Which reduces to the surprisingly simple expression: u = rac {C + 1} {C + 2} Replacing our substitutions yields the angle that produces the maximum range for uneven ground, ignoring air resistance: heta = rccos left( sqrt { rac {2 g y_0 + v^2} {2 g y_0 + 2 v^2} } ight) or equivalently heta = rcsin left( rac {1} {sqrt {2 left(1 + rac{g y_0}{v^2} ight) }} ight) Note that for zero initial height, the elevation angle that produces maximum range is 45 degrees, as expected. For positive initial heights, the elevation angle is below 45 degrees, and for negative initial heights (bounded below by y_0 > -0.5 v^2 / g ), the elevation angle is greater than 45 degrees. The maximum range achieved at the optimum angle is found to be R_{ m max} = rac{v}{g} , sqrt{v^2 + 2 g y_0} Example: For the values g = 9.81 , { m m/s}^2 , y_0 = 40 , { m m} , and v = 50 , { m m/s} , an elevation angle heta pprox 41.1^circ produces a maximum range of R_{ m max} pprox 292.1 , { m m} . Angle of impact The angle ? at which the projectile lands is given by: an psi = rac {-v_y(t_d)} {v_x(t_d)} = rac {sqrt { v^2 sin^2 heta + 2 g y_0 }} { v cos heta} For maximum range, this results in the following equation: an^2 psi = rac { 2 g y_0 + v^2 } { v^2 } = C+1 Rewriting the original solution for ?, we get: an^2 heta = rac { 1 - cos^2 heta } { cos^2 heta } = rac { v^2 } { 2 g y_0 + v^2 } = rac { 1 } { C + 1 } Multiplying with the equation for (tan ?)^2 gives: an^2 psi , an^2 heta = rac { 2 g y_0 + v^2 } { v^2 } rac { v^2 } { 2 g y_0 + v^2 } = 1 Because of the trigonometric identity an ( heta + psi) = rac { an heta + an psi } { 1 - an heta an psi } , this means that ? + ? must be 90 degrees. Actual projectile motion In addition to air resistance, which slows a projectile and reduces its range, many other factors also have to be accounted for when actual projectile motion is considered. Projectile characteristics Generally speaking, a projectile with greater volume faces greater air resistance, reducing the range of the projectile. This can be modified by the projectile shape: a tall and wide, but short projectile will face greater air resistance than a low and narrow, but long, projectile of the same volume. The surface of the projectile also must be considered: a smooth projectile will face less air resistance than a rough-surfaced one, and irregularities on the surface of a projectile may change its trajectory if they create more drag on one side of the projectile than on the other. Mass also becomes important, as a more massive projectile will have more kinetic energy, and will thus be less affected by air resistance. The distribution of mass within the projectile can also be important, as an unevenly weighted projectile may spin undesirably, causing irregularities in its trajectory due to the magnus effect. If a projectile is given rotation along its axes of travel, irregularities in the projectile's shape and weight distribution tend to be canceled out. See rifling for a greater explanation. Firearm barrels For projectiles that are launched by firearms and artillery, the nature of the gun's barrel is also important. Longer barrels allow more of the propellant's energy to be given to the projectile, yielding greater range. Rifling, while it may not increase the average (arithmetic mean) range of many shots from the same gun, will increase the accuracy and precision of the gun.

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