How do you solve for theta in the range equation for launched projectiles?
109151
It looks like your equation got cut off, but I think you were writing\n\nR = (V^2/2g)(sin (2 theta) +sin (2 theta))\n\nwhich can be immediately simplified, by adding the sin (2 theta) terms, as \n\nR= (V^2/2g )(2)(sin(2 theta) = (V^2/g )sin (2 theta)\n\nwhich is the formula that the other correspondent already gave.  To solve it, simply multiply both sides by g, divide by V^2, and you get \n\n(R g)/v^2 = sin (2 theta)\n\n[Arcsin (Rg/V^2) ]/2 = theta
730340
Projectile range would be the equivalent of horizontal distance.\nThe forces involved are gravity (acting on the projectile's mass in the vertical direction), and the "push" from the initial thrust.\n\nAs the angle of the firing position increases above the horizontal, more of the initial push is used to combat gravity.\n\nIf the angle is too low, the projectile will meet the ground before it's maximum range (not enough action against gravity).\n\nFor most projectiles, the optimum angle is between 42 and 45 degrees (depending on the strength of the initial impetus). This is because the trajectory approximates a parabola.\n\nWith the given choices, the answer will have to be 45 degrees (answer B)...
470748
When it hits the ground h = 0 \nHence the equation reduces to\n16t^2 - 70t -3 = 0\nt = (70 plus or minus sqrt(70^2 - 4*16*(-3)))/2*16\n  = (70 plus or minus sqrt(5092))/32\n  = (70 plus or minus 71.35)/32\nt= 4.42 s approx  (note that 70 - 71.35 is not applicable since it is negative)\nAnswer 4.42s
966696
Arc=r.theta=8*pi/8=pi
754389
Learn to answer these questions yourself!\n\nIf the air is assumed to be frictionless, the vertical and horizontal velocities are independent.  The altitude is given by H = H0 + V0t + 1/2 a t^2.  H0 is +30 meters, V0 is +24.5 m/sec, and a = -9.8 m/sec^2.  If the target is at ground level, you are solving a quadratic for H = 0.\n\nSolving the quadratic gives you the time of flight, t.  Multiply the time of flight by the horizontal velocity (49 m/sec * cos(30°)) and  you get the distance of the target.\n\nOf course, you can just look at the problem and the answers and see if there's just one answer that's anywhere close.  In this case, the initial vertical speed is 24.5 m/sec; it takes 2.5 seconds for the arrow to rise to its peak altitude, and another 2.5 seconds to come back down to the altitude of the bow (5 seconds total).  Falling the final 30 meters would take on the order of a second.  The cosine of 30 degrees is about 0.87, so the horizontal velocity is about 42.5 m/sec; multiply by 6 seconds, and you get about 250 meters.  Ergo, (4) is the only answer that's even close.
1084130
The projectile motion is the motion of a projector starting from origin (x=0 & y=0) with initial velocity (v) & inclination (a) where (a) is the angle between X-axis & (v) so the only accelerations affecting the projector are:\nx''=0 & y''=-g --------> (1) & (2)\n\nTo calculate velocity (x' & y'):\nfrom (1) & (2) we get:       \nx''=dx'/dt=0  \ntherfore: x'=constant=v cos(a) [from initial condition] --------> (3)\n \nfrom (3) we get: y''=dy'/dt=-g\ntherfore: y'=-gt+constant\n\nfrom initial condition:\nat t=0 ----> y'=v sin(a) & constant=0\ntherfore: y'=v sin(a)-gt ------> (4)\n\nThe parametric equation:\nfrom (3): dx/dt=v cos(a)\ntherfore: x=tv cos(a)+constant\n\nfrom initial condition:\nconstant=0\ntherfore: x=tv cos(a) -----> (5)\n\nfrom (4): dy/dt=v sin(a)-gt\ntherfore: y=tv sin(a)-0.5g(t^2)+constant\n\nfrom initial condition:\nt=0 therfore y=0 & constant=0\ntherfore y=tv sin(a)-0.5g(t^2) -----> (6)\n\nfrom (5): t=x/[v cos(a)]          sub in (6)\ntherfore\ny=[x tan(a)]-{ [g(x^2)]/[2(v^2)] }*[1+ tan^2 (a)] \nthis equation is called The Cartizian Equation\n\nTo find The Range ------> sub y=0 in The Cartizian Equation\ntherfor:\nR(range)=[2v sin(a) cos(a)]/g
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
793793
the thing this problem is concerned is the vertical component of the projectile.\n\nsince the dart is thrown horizontally, the initial vertical velocity is zero.\n\nthe vertical velocity of the dart just before it hits the dart board is\n\nv^2 = u^2 + 2gs\nv^2 = 0 + 2(9.8)(0.32)\nv^2 = 6.272\nv = 2.50 m/s\n\nthe time it took the dart to hit the dart board is given by\n\nv = u + gt\n2.50 = 0 + 9.8t\nt = 0.255 sec\n\n\nnow, we have the time of travel of the dart, we can use the formula  d = vt  to find the horizontal distance\n(the horizontal velocity remained constant throughout the flight)\n\nd = vt\nd = 12.4m/s * 0.255s\nd = 3.169 m\n\n\ntherefore the dart player is standing  3.169 meters away from the board...\n\nthere ya go...
703199
It is an exercise in elementary calculus to show that (ignoring air resistance) the elevation angle for maximum distance is 45 degrees.  The vertical velocity of the egg will be zero at the peak height, and will increase as v = gt^2/2 until it hits the ground.  Energy is conserved, so the vertical velocity when it strikes the ground will match the vertical velocity with which it was launched, and the horizontal velocity will be constant.   This is enough data to solve the whole thing, but there is a shortcut: in the parabolic path at issue, it is always true that the peak height is 1/4 of the horizontal distance.  Hence, the peak height here is 24.225 meters, and that makes it trivial to compute the time and hence the velocity.
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
610143
I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
199351
I haven't worked with this in a long time and I'm not sure this will work, but try starting w/ this:\n1) Define an equation that describes the " ∞ " with polar coordinates (something like r = C1 - C2*theta);\n2) If L is lengh, delta.L = delta.theta / (2 pi) * r\nand dL = r/(2 pi) dtheta  ;\n3) substitute for r and integrate from 0 to 2pi (or the necessary for one complete " ∞ ".\n \nProbably you'll find more than one mistakes here, but I think the general idea is correct.
965298
Maximum range is reached if the missile is launched at an angle of 45 degrees from the horizontal. This can be proved using calculus. Thus the problem can be solved using this known fact and the given range of 8500km.\n\nThe vertical component of the velocity =vsin45. Since the angle is equal to 45 degrees, the horizontal component is the same, i.e. vsin45.\n\nThe time it takes the missile to travel 8500km  can be computed using the basic formula: time=distance/velocity. Thus \n\nt =8500/vsin45  Eq. 1). Where vsin45 is the horizontal component of the initial velocity of the missile.\n\nNow use the formula: v^2-u^2=2as, and substitute known values, as follows:\n \n(vsin45)^2-0=2*9.8s Equation 2 \n\nWe take vsin45 as the final vertical component of the velocity; u=0, being the initial velocity at the top of the flight of the missile; a the acceleration of gravity equal to 9.8m/s^2, and s is the maximum altitude reached in the trajectory.\n\n(You also have to know that the initial vertical component of the velocity upward is equal to the final vertical velocity downward  of the free falling body dropped from the top of the flight of the missile.)\n\nWe now use another formula for s:\n  \ns=ut+1/2at^2\n  s=0+1/2*9.8t^2   Equation 3)\n\nBut from equaton 1) t=8500/vsin45 This is for the entire flight of the missile. The time for a free falling body to reach the ground from the top of the light of the missile is only half this. Thus divide the given expression by 2 and substitute this in  equation 3):\n\ns=1/2*9.8*[1/2*8500/v sin45]^2\n\nsubstitute the value of (vsin45)^2 given in equation 2):\n\ns=1/2*9.8*1/4*8500^2/2*9.8s\ns=1/16*72250000/s\ns^2=4,515,625   \ns=2,125 km
927920
The distance is V²/2a where V = muzzle velocity and a = g(sinθ + μcosθ) where θ = angle above hoizontal of the incline and μ = coefficient of friction.\n\nNote that when θ is <0, a can be 0 and the distance is infinite, as it never slows down at all.
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec
1135263
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.