A spacecraft is traveling with a velocity of v0x = 4750 m/s along the +x direction.?
98111
Code:\nx'' - 2nd derivative of x with respect to time (t) (acceleration in x)\n\nx' - 1st derivative of x wrt time (velocity in x)\n\nAnd the same system for the y coordinate.\n\nx''=1.2 then by integration wrt time gives x'=1.2t+constant. Boundary conditions are x'=4750 m/s at t=0, so the constant = 4750 m/s. Equation is now Vx = x' = 1.2t + 4750. Substitute in t=743s, answer is Vx=5641 m/s. \n\nSame method for y'' with acceleration of 8.5 m/s^2, and boundary conditions of y'=0 at t=0 gives Vy=6315.5 m/s.\n\nFinal answers:\nVx=5641.6 m/s\nVy=6315.5 m/s
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
949041
This problem relies on Newton's second law: F=ma\n\na. This question is not completely specified.  The mass of the rocket is then (F/a=m or 2.0E7/a).\n\nb. We first find the net force on the rocket and use the mass from the previous question. Now we compute the acceleration at liftoff (F/m=a or (25E6 - 2.0E7)/(2.0E7/a)).\n\nc. We derive a new formula by double integration on acceleration. \nintegrate(a dt) = at, integrate(at dt) = .5at^2. The sequence gives us acceleration -> speed -> displacement at time t. The rocket will be at velocity: +(25E6 - 2.0E7)/(2.0E7/a)(7*60) m/s and at displacement of: +.5((25E6 - 2.0E7)/(2.0E7/a))(7*60)^2 m.
59053
actually... although the first answerer did the correct calculation, he left out a step. that is the acceleration produced by the rocket, however you must take gravity into account. \nF=ma, a=F/m= 23.3 is the acceleration produced by the thrust of the rocket, but the rocket is not actually going to accelerate that much because an acceleration of 9.8(due to gravity) is pulling it down in the opposite direction. so 23.3-9.8= 13.53\nso the initial acceleration of the rocket is:\n13.53m/s^2
798759
1- yes, 2-yes, 3-no Vx=the same 24.5m/s, 4-yes, 5- no Dy=21, 6-yes Ax=0 throughout the whole parabola. 7- yes coz the only force (constant force) is gravity. Now (8) – its final Vy=26, and Vx=  still the same 24.5, thus Vtotal=sqrt(Vx*Vx+Vy*Vy)
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
726547
Find the horizontal and vertical components of its velocity.  (Hint: one is v*sin(ø) and the other is v*cos(ø).)   Each velocity times the ariplane's mass is that direction's component of momentum.
772033
Draw a picture of it with arrows representing the vertical and horizontal velocities as 2 legs of a right triangle.  This triangle will show you that the angle with the vertical is arctan 3.5/5.5 = theta\n\nThen the flight path V will be Vv/cos(theta) or Vh/sin(theta)\n\nSince you're ruling out pythagoras, you really need to find the angle first, then the speed.......
883093
speed = 55km/h = 275/18 m/s\ndistance 250 + 520 = 770m\ntime = distance/speed\n= 770*18/275 Seconds\n= 50.4 seconds
196152
Terminal velocity. Next!
958243
You need to know the initial speed.  Then the answer is:\n\n(v - u)/t\n\nwhere u = initial speed\n           v = 0 (final velocity)\n           t = 5 seconds\n\nThe answer will be negative because it is decelerating
716722
For launch velocity\nv0 = 4.47 m/s,\nlaunch angle\nθ = 66 degrees,\nThe horizontal range is\nR = 1.515173718712374 m.\nThe total time of flight is\nt = 0.83337718278825 s.\nThe peak height is\nh = 0.8507839727702984 m.
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
392981
Is the speed at which it travels.
782512
You need to use kinematics and Newton's 2nd law to solve the problem.  In the horizontal direction, Newton's 2nd law states:\n\nForce - friction = m*a.  Since you are given the values, you can solve for the accleration:\n\n1149 - 909 = (2700)*a, solving for a, you get:\n\na = 0.089 m/s^2\n\n----------------\n\nNow use kinematics to solve for the distance needed to reach the desired final velocity.  Use the following kinematic equation:\n\n2ax = v_f^2 - v_i^2.  You know that:\n\nv_f = 3.2 m/s\nv_i = 0 m/s (starting from rest)\na = 0.089 m/s^2\n\nNiw solve for x:\n\nx = [v_f^2 - v_i^2]/(2a) = 3.2^2/(2*0.089) = 57.6 meters\n\n---------------\n\nHope this helps
749570
V = √(550² + 80²)\n≈ 556 mph\nDirection tanθ = 550/80 in 3rd quadrant\nθ ≈  81° 43' 27" from West\n= 188° 16' 33" T (clockwise from N)
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....