Velocity Physics Question?
368641
I'm not sure what 500 above the horizontal means - radians maybe?  Or is there something missing, like a launch angle, and that you're 500 ft above ground when the ball is thrown?\n\nIf I'm picturing the scenario right, you're basically shooting a ball up at an angle and finding all of the specifics - max height, velocity at impact, etc.  All of those can be found using some of the following equations:\n\nvf^2 = vi^2 + 2ad\nvf = vi + at\nd = vt\n\nwhere\nvf = final velocity\nvi = initial velocity\na = acceleration due to gravity (9.8 m/s^2 or 32.2 ft/s^2)\nt = time\nd = distance\n\nHorizontal and vertical velocity are independent of each other.  The vertical velocity changes due to gravity, and the horizontal velocity will remain constant throughout the ball's flight (assuming no air resistance is in play).  So to find the max height, we'd need the vertical velocity component of the 110 ft/s.  That comes out to 110 sin x (where x is that launch angle I asked for earlier).  So this will be your vi in your first equation.  vf = 0 at the apex (the ball has stopped and is about to reverse direction), and since you are decelerating, use -32.2 ft/s^2 (I know you specified 9.8 m/s^2, but the units don't match the 110 ft/s - you can always convert one or the other if you like).  Solving that first equation will give you the distance it traveled.  If there is an additional height involved (remember if I asked if you were 500 ft above ground?) then your total height from ground level would be your answer from the equation plus 500.\n\nThe time to the apex can be found with the second equation.  Again, using vi = 110 sin x and vf = 0, you can solve for the time (and again, keep your units consistent or else the answer doesn't make sense).\n\nLet's jump to the last question about the velocity of the ball upon impact.  If it falls to the same height that it was thrown at, then the velocity on impact will be the same as the launch velocity - 110 ft/s.  But if there is that additional 500 ft height, we can use that first equation to find the vertical component of the impact velocity.  If we were to picture that part of the equation as starting from the apex, vi = 0.  The distance d will be the distance found in part (i) plus 500 ft.  So you can solve for the final velocity there - but remember that this only solves for that vertical component of the total velocity.  You'd need to find that total via the Pythagorean theorem - picture a triangle with the vertical velocity as one leg, and 100 cos x (the horizontal component of the original launch velocity) as the other leg.  Square both, add, and take the square root to find the total velocity on impact.\n\nOn to part (iii) - now that you have the final velocity (just use the vertical component as calculated above), you can use the second equation to find the time it took to fall from the apex to the ground.  Add this to the answer in part (ii) and you have the total time in flight.\n\nDown range, I'm assuming, is the horizontal distance the ball traveled.  Using the third equation, you can find that.  The velocity in this case is the horizontal component of the launch velocity - 100 cos x.  Take this times the time from part (iii) and you have the range.\n\nHope this helps - figure out what that launch angle is and how that 500 plays in to the equation and you should be on your way!
1045430
Momentum - P = mv (P=momentum, m=mass, v = velocity)\nThree Laws of Newton - F = ma (F=Force; a=acceleration)\nEquations of motion - 1) v = u + at\n                                    2) S = (1/2)ut + at^2\n                                    3) v^2 = u^2 + 2aS\n(v=final velocity; u = initial velocity; S=distance)\n\nIf this can help you please mail me back !\nBEST OF LUCK !!!!
340116
a) d= 0.5 * 9.8 * t^2 + Vi * t = 10 m\nFor t=1 sec:\nVi= 10 - 9.8*0.5 = 5.1 m/s \n\nb) g= (Vf-Vi)/ t\nFor t=1sec and Vi=5.1\nVf= 9.8+5.1= 14.9 m/s\n\nHope it is clear enough.
999681
I'm assuming you meant that 20m down, the two objects collide. Right?\n\nFor the object that is falling down, you know:\n\na = 9.8 m/s^2\nd = 20m\nVi = 0 m/s^2\nVf = ?\nt = ?\n\nUsing the formula: Vf^2 = Vi^2 + 2ad\n\nYou can find that: \n\nVf = 19.798989....m/s^2\n\nYou plug that into the next formula to find time (t), which is: \nVf = Vi + at\n\nYour answer should be t = 2.020305......s\n\nYou use that number for the second object following the formula: d = vt (because the horizontal velocity will never change since there is NO horizontal acceleration)\n\nd = 2.5(2.0203....)\n\n= 5.05m\n\nI hope there aren't any mistakes! ;D
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.
716223
The change in velocity of an object undergoing uniform acceleration, a, for a given time, t, is given by:\nv = at\nwhere v is the change in velocity of the object.\n\nIf the object is accelerating at a constant rate of 10 m/s^2 for 5 seconds, the change in velocity the object experiences is 50 m/s.\nSince (we assume) the object started from rest, the objects final velocity after 5 seconds is 50 m/s.\n\nTo find the average velocity, divide the total displacement of the object by the total time.\nIn this case, the object has traveled 125 meters in the 5 seconds it has been accelerating (d = 1/2 at^2), so dividing this by 5 seconds and we get an average velocity of 25 m/s.
735384
Using the law of conservation of energy determin\n\na) the vertical component of vector V(initial)\n\nIt went up 20 meters, mgh = 1/2mV^2, solve for velocity\n\nsqrt[2(9.8m/sec^2*20meters)] = v = 19.8m/sec\n\nb) the work done by the gravitational force on the particle during its motion from P to the ground.\n\nMaybe this is a trick question, I will take it at face value, when at P it is 60meters above the ground, the work of the gravity bringing it down from P is just mgh = 0.5kg*9.8m/s^2*20m=98J\n\nc) the horizontal and vertical components of the velocity vector when the particle reaches the ground:\n\nHorizontal velocity is constant at 30m/sec, nothing to calculate.\n\nVerticle velocity, keep in mind that the object reached its maximum height at 80 meters high (ie from a verticle perspective is is at zero velocity at that point, and begins to fall, the terminal vertical velocity comes from the potential energy of \nmgh = 0.5kg*9.8m/s^2*80m = 1/2 mV^2, solve for v\n\nvertical velocity is 39.6 m/sec downward\n\nplease check for calculation errors.
14955
(a) it's an inelastic collision.  An inelastic collision is when they stick together, and an elastic collision is when they bounce off of each other\n(b) p=mv  momentum= mass x velocity.  p= 30 m/s x 500kg = 15,000 kg*m/s\n(c) well, momentum is conserved, so the we would use the formula from before and the momentum(p) from before, except that now the mass is different because it is the combined masses of the two vehicles (300+500=800) , so you can solve for the new velocity.\np=mv  15000 = 800 x v\n\ndivide both sides by 800 to get  v=15000/800 = 18.57 m/s
495016
This is more of a physics question...\nd final = .5*a*t^2 + v*t + d\nv final = a*t + v\nd is displacement, v is velocity, a is acceleration.\nEnjoy.
357903
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
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
757769
Here's how to solve this.\nThe velocity is a vector, and right-angle components of velocity operate independently.\nThe bullet is fired with 2000 ft/s horizontal velocity but zero vertical velocity. So you know the horizontal and vertical components of velocity at the instant the bullet leaves the barrel.\n\nHowever, once it leaves the gun's barrel, the bullet experiences no horizontal forces, but vertically it is acted upon by gravity.\n\nSo: 1) how long does it take for the bullet to fall 5 ft, starting with zero vertical velocity? (hint: s = 1/2 a t^2\n\n2) Once you figure out how long until the bullet hits the ground, then figure out how far the bullet has travelled horizontally during that time, given 2000 m/s initial horizontal velocity and no horizontal forces.\n\nDoes this help?
200570
the velocity of the ambient
813594
v = wave length * f.\n\nSo, length = v / f =   1,7m
946809
x=v(horizontal velocity of water hitting ground)*t( time of travel)\n  = sqrt(2gy) *t; from equation v^2=2gy\nt= sqrt(2h/g); from equation h=.5*g*t^2\nthere you get that equation. I did with kinematics.
107382
v^2-u^2 = 2gh\nHere, v=1, g=-10m/s/s, h=0.6m\nHence, u^2 = 1+12=13\n=> u = 3.606 m/s
495603
Assuming that the energy of the ball remains constant (i.e. no air resistance), the speed will be the same. The VELOCITY, however, will be -35 m/s.\nSpeed and velocity are NOT the same!