in gta vice city how can i find the plus + speed with rockets propeler (fire from behind vehicels) then flying
595750
YOUR IN THE WRONG PLACE TO BE ASKING THAT
841719
sorry for laughing but I know everything can be used with math. Those rockets, how long is goes up, where it lands the wind , angle bla/bla
769161
Its a pain to do this with X and Y components, but there is an easier way....\n\nRange Formula - used when start and end heights are the same\nQ = Theta\nd = ((Vo^2)(sin 2Q))/g\n\n301.5 = ((Vo^2)(sin 2(25)))/9.81\n\nVo = 62.13 m/s\n\nNow that u have initial velocity, find the y component\n\nVo Sin Q\n\n62.13 sin 25 = 26.26 m/s\n\nThen No Time Formula\n\nVf^2 = Vo^2 + 2gd\n\n0 = (26.26^2) + 2(-9.81)d\n\nd = 35.15 m\n\nHope this helps
79521
It depends on the type of bullets and the type of gun that you use to fire the bullet.\n\n.22 rimfire cartridge 1200 - 1500 fps \n.22 centerfire cartridge 2400 - 3000 fps \n.22 Swift 4000 fps \n.38 Special 600 fps \n.221 Fireball 2650 fps \n\nThis gives you an idea of the velocity of bullets.
157080
This is just the regular projectile motion problem (see your text) just with an added initial velocity in the x direction (the velocity of the ship).\n\nLet's take as our positive direction the direction the gun shoots. (That is, the ship is moving in the negative direction.) Thus, the initial x component of velocity of the projectile is the x-component of the guns velocity less the velocity of the ship. (I'm going to use Vo for the initial velocity, since using "not" as a subscript confuses me.)  And I'll use A for alpha to save space.\n\nVxo = Vo cos(A)- v\nVyo = Vo sin(A)\n\nOur equations of motion are\n\nX = Xo + Vox t + 1/2 ax t^2\nY = Yo + Voy t + 1/2 ay t^2\n\nTo find the range, we want when the projectile splashes back down again, i.e. Y = 0.  If we set Xo = 0, then X will be the range, i.e. the X position when Y = 0 again.  And there is no acceleration in the x direction, the y acceleration is -g, the acceleration of gravity.  Plugging all that in, we get\n\nR = Vxo t \nR = (Vo cos(A) - V) t                     (1)\n\n0 = Vyo t - g t^2/2\nVyo t = g t^2/2\nVyo = g t/2\n2 Vyo/g = t\n2 ( Vo sin(A)) / g = t                     (2)\n\nPlugging our expression for t (Eqn. 2) into Eqn. 1 yields\n\nR = (Vo cos(A) - V)2 ( Vo sin(A)) / g \n\nwhich is the formula you gave.\n\nTo find the max range, we can use calculus.  There may be a more straightforward way if you don't know calculus, but this works for me.\n\nThe range formula is a function R of variable A.  To find the max, take the derivative and set it equal to 0.  Thus\n\ndR/dA = 2 Vo/g [sin(A)(-Vo sin(A)) + (Vo cos(A) - V)cos(A)]\n       =2 Vo/g [-Vo sin^2(A) + Vo cos^2(A) - V cos(A)]\n\nusing a trig identity\n\n       =2 Vo/g[Vo cos^2(A) - Vo + Vo cos^2(A) - V cos(A)]\n       =2 Vo/g[2Vo cos^2(A) - V cos(A) - Vo]\n\nWhat we have here is a quadratic equation with the variable cos(A).  When we set it equal to 0, in order to find the minimum, we can use the quadratic formula.\n\ncos(A) = (- (-V) + SQRT[V^2 - 4(2Vo)(-Vo)])/2 (2Vo)\n        = (V + SQRT[V^2 + 8Vo^2])/4Vo
909410
Strange formula, since it has no consideration of the changes when the engine runs out of fuel.\n\nSuggestion, if you have Excel or something like that...\n\nPlug in a series of times in one column - perhaps seconds, or for more precision, try tenths of seconds.\n\nHave the next column calculate the height based on your formula.\n\nThen you could calculate vertical speed in the next column.\nYou can also have Excel plot graphs of this....\n\nSounds like your homework is the usual kind.  Maybe this will help.  \n\nClearly, if you differentiate the expression, you can get a formula for vertical velocity (dH/dT)  \n\nIf you solve for t when vertical velocity = 0, that will be the time of maximum altitude...\n\nAnd so on.  Good luck.
1119837
First divide up the initial velocity into its horizontal and vertical components.\nV is the initial velocity of 40 m/s.\nTheta is the launch angle of 37 degrees above the horizon.\nV_x_i and V_y_i are the horizontal and vertical components of the initial velocity respectively.\n\nV_x_i = V * cos(theta) = 40 * cos (37) = 31.95 m/s\nV_y_i = V * sin(theta) = 40 * sin (37) = 24.07 m/s\n\nAlthough the vertical component of the object's velocity will change as the object experiences gravitational acceleration toward the ground, neglecting air resistance, the horizontal component of velocity can be treated as a constant.\n\nV_x = V_x_i = 31.95 m/s\nV_y = V_y_i  - g*t\nwhere t is the time after the object is launched and g is the gravitational acceleration experienced by the object (g = 9.81 m/s^2).\n\nIf the fence which the object hits is 128 meters away (in the +X direction), then the object (traveling at 31.95 m/s in the +X direction) will hit it at about 4.01 seconds after launch.\n\nBy taking this value (4.01 s) and plugging it in for t in the below shown equation to solve for the distance relative to the initial launch height the object has traveled,\nd = V_y_i * t + 1/2 (g)t^2\none can calculate that the object is at a height of about 17.65 meters above its original launch height.\n\nSo I calculate the impact height to be 17.65 meters.
239661
The bullets won't be pulled to earth at the same downward _speed_, but they will be pulled by the same downward _force_.  The force that pulls it toward the earth (gravity) would cause both to have the same vertical speed, assuming:\n\n(1) the bullet that was shot from a gun is fired horizontally, and\n(2) these occur in a vaccuum, where the effects of air can be neglected.\n\nNow, to find how long it would take them to hit the ground and how far away the fired bullet would be, you'd need to know how high the bullets are initially and how fast the bullet came out of the gun.  There are a few formulas that you can use:\n\nvf = vi + at\nvf^2 = vi^2 + 2ad\nd = vt\n\nwhere\nvf = final velocity\nvi = initial velocity\na = acceleration due to gravity\nt = time\nd = distance\n\nUsing these in a logical manner can get you what you need.  For example, you can use the second equation to find the vertical velocity of the bullet as it hits the ground.  Then you can take this vertical velocity (vf now) and use the first equation to find how long it took to fall.  Then with this time, you can find out how far the fired bullet went using the third equation.\n\nIt helps if you separate the bullets' velocity into a vertical and horizontal component - i.e. how fast is it moving horizontally and how fast vertically.  These two are independent of each other.  If you keep that in mind that helps in understanding why, in a vaccuum anyway, these two bullets should hit the ground at the same time.
566525
let V1= 4 mph, V2= 5 mph\n\nV1*t + V2*t = 3\nt= 20 min\n\n19. a 20 min\n\n\n14.16/3 = 4.72\n\n4.72*14 = 66.08\n\n20 c 66.08
151131
well i have a raptor 660 and the way i got it fast were simple things. complete exhaust kit and intake, jetting your carb is also a very good way to make it faster, i have all that done to my raptor and i got a 30 mph gain out of it all
861574
first find the centripital acceleration\na_c = v^2 / r\na_c = 140^2/1000\na_c = 19.6\n\nthe total acceleration at the bottom of the loop is:\na = a_c + g\na = 19.6 + 9.8\na = 29.4\n\nF = m*a\nF = 70*29.4\nF = 2059 Newtons\n\n29.4/9.8 = 3 so the pilot is pulling 3 Gs at the bottom of the loop
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
861581
Cops have them too.  Use helicopters.  Require certain high speed models to come with a kill switch like are used on go-cart tracks.  Disabling it would activate a lo-jac type device or ruin the engine.
735664
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
731455
Something that may be fun is a project with model rockets. Give them the necessary information to calculate the height of the rocket, based on the rocket engine and mass of the rocket as homework.\n\nThen at school, fire a few of them and using basic trig calculate the height and see how close they are.\n\nRocket engines are typically rate something like: A8-3 which breaks down as follows:\nA --> 2.50 Newton-seconds of power\n8 --> average thrust (in newtons)\n3 --> seconds from end of thrust to ejection charge (parachute)\n\nThe engines also have a time-thrust curve that shows how the power is delivered over time. Check the following link below for some more info:\n\n------------------------------------------------------------------------\n\nHere is another one:\nNot sure how readily available props are for you but if you can hook up a camera in a room with a strobe light/flash and then set up a collision (air hockey table or pool table come to mind) then you can calculate the velocity of each object, measure its mass and then demonstrate the conservation of momentum using vectors. Because you know the frequency of the strobe, and the distance traveled, you can compute the vector velocity.\n\nOr how about:\nFirst, fire an arrow down a football field. Given the amount of time aloft calculate the initial velocity (ignore air resistance/friction) Then fire an arrow at a bale of hay (extend the bow by the same amount) and measure how far into the bale the arrow went. Using the same initial velocity and distance into bale, calculate the rate of deceleration and coefficient of friction of the hay bale.
958247
v=v0-at=0=96-32t\n32t=96\nt=3sec\n\na=32'/s^2\nv0=96'/s\nd=5+96t-16t^2\nd=5+96*3-16*9=5+288-144=5+144=149'\n\nA) The arrow will reach a height of 149'\nit takes 3sec to go up & 3 more to come down to 5'\nthe last 5'\n5=96t+16t^2\n16t^2+96t-5=0\nt=(-96+/-√96^2+4*5*16)/32\nt=-3+/-√9216+320=-3+√9536/32\nt=-3+97.65/32=.05\nB) total time=3+3+.05=6.05sec
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc