A physics question: when firing a canon ball what is the total time of flight and the total horizontal range?
338055
Ok, separate it into the verticle and horizontal aspects.  The verticle aspect will tell you how long the cannonball is in the air, since you will be able to figure out what the verticle velocity of the cannonball is.  It will have an acceleration of -9.8 meters/sec^2 upwards until it reaches its highest point, at which time it will stop and then have an acceleration of 9.8 meters/sec^2 downward (aka it will be in free fall) until it hits the ground.  Knowing this, you should be able to figure out how long it will be in the air.\nOnce you know how long its in the air, you can calculate the distance it will go.  The horozontal aspect of its original velocity will be the velocity forward.  This will not change so take this velocity (m/s) times the time it will be in the air (s) and you will get the distance the cannonball will go (m).
368139
If the object has onboard power, then the speed is not really relevant to whether it will leave the atmosphere, just when. If you just fire an object upwards and it has no self-propulsion, however, then you need to calculate the change in gravitational potential energy (find out how far the atmosphere extends) and add the energy lost to air friction (this depends on velocity). Equate this to the object's initial kinetic energy and solve for initial velocity v. Your initial velocity will have to be larger than this.
141479
You should do your own homework problems.  Being told answers will not help you learn.\n\nI'll try to help you understand, though.\n\nThink about this.  The tiger's speed accelerates at a rate of 9.81 m/s^2.  For constant acceleration, distance is related to time by the equation d=t*v_0+(t^2)*a where v_0 is the initial speed IN THE DIRECTION OF ACCELERATION, d is the distance traveled, a is the rate of acceleration and t is the time.  Use that to figure out how much time it takes the tiger to reach the ground.  Then multiply this amount of time by the forward speed.  This tells you how far the tiger has moved forward upon landing.\n\nFor the ball the same formula applies.  It works for the projectile too, but you must decompose the velocity into its vertical and horizontal components (hint:  one will be the sine of the angle it's fired from times the speed and the other will be the cosine of that angle times the speed).  I'm going to leave it up to you to figure out which.  This question obviously has more to do with wanting your homework answered for you than with personal curiosity, because of its specificity, and if you just didn't get the concepts it's ok.  It is not ok to submit your homework on the internet in hopes that others will do it for you.
878067
As long as neither of the balls starts spinning (or changes the rate at which it's already spinning) after the collision (in which case some of their linear momentum would have been converted to angular momentum), the total linear momentum of the system before and after the collision must be the same.\n\nLet p11 and p21 be the momentum of balls 1 and 2 before the collision; let p12 and p22 be the momenum of balls 1 and 2 after ths collision; and let m be the mass of the balls (which we are told have the same mass).  The momenta are all vector quantities, which I'll write as order pairs of numbers (i.e.,  p = (x,y) = x*i + y*j, where i and j are the unit vectors in the +x and +y directions, respectively).\n\nLet the +x axis be pointed in the east direction, and the +y axis be in the north direction.\n\nThen, writing each \n\np11 = m*(2 m/s, 0 m/s)\np21 = m*(-1.2 m/s, 0 m/s)\np22 = m*(0 m/s, 1.43 m/s)\np12 = m*v12 (v12 is the vector velocity of ball 1 after the collision)\n\nBy conservation of momentum, we have that:\n\np11 + p21 = p12 + p22\n\nm*(2 m/s, 0 m/s) + m*(-1.2 m/s, 0 m/s) = m*v12 + m*(0 m/s, 1.43 m/s)\n\nv12 = ((2 - 1.2 - 0) m/s, (0 + 0 -1.43) m/s)\n\nv12 = (0.8 m/s, -1.43 m/s)\n\nThe speed of ball 1 after the collision is given by:\n\nsqrt((0.8 m/s)^2 + (-1.43 m/s)^2) = 1.639 m/s\n\nThe angle the velocity vector of ball 1 makes with respect to the +x axis after the collision is given by:\n\narctan(-1.43/0.8) = -1.061 radians = -60.776 degrees\n\n(It helps to draw a picture of this to visualize why this relationship holds.)\n\nThe first ball is traveling 60.776 degrees south of east after the collision.
1102976
a) It will decrease in speed as soon as the accelerative force of the explosive gas ceases to be greater than the drag of the gun barrel or the air or other media it passes through. This could mean that it begins to slow down in the barrel, or soon after leaving it. Certainly, after a few feet from the gun barrel one would expect the gases from the gun to be dissipated and the bullet to begin to be dragged by the air. However, the reduction in speed of a high velocity bullet could be negligible for seconds after firing.\n\nb) As soon as the bullet leaves the barrel it will begin to drop towards the earth. The formula for this is: \ndistance =0.5*acceleration*(time-squared)\n\nAcceleration at sea level is 9.81 metres per second per second. Thus it will take almost exactly 2 seconds for the bullet to drop the 20 metres from your firing platform. (square root of [20/(0.5*9.81)]  )\n\nc) See b above. The bullet is acted upon by gravity at all times. It is only prevented from being accelerated downwards by a force equal or greater to its weight pushing it up. In your example the gun barrel supports it until the bullet leaves the barrel, then begins to fall. Even if you pointed the gun straight up, the bullet will be slowed by gravity as soon as it leaves the barrel and gases behind, gradually slowing to a stop and falling back to land on your head.\n\nIf the bullet had enough horizontal velocity to make it to the horizon within two seconds, it would have acheived the required "escape velocity" to leave or orbit the earth, always falling but never landing. As the horizon at sea level is eight miles away this would need a speed of 28,800 miles per hour or 46,340 kilometers per hour. \n\nThe speed of sound is c.1,220 km/hour, so this would be 38 times the speed of sound!\nThe muzzle velocity of most military calibres is c.700metres/second or 2, 520 km/h or twice the speed of sound.\n\nThus in the two seconds it takes the bullet to fall to earth it would have carried some 1,400 metres.\n\nd) Very much so!
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet
654739
The horizontal position of the projectile is described by x(t) = (v_x)*t + h since the horizontal force is zero. v_x is the initial horizontal velocity and h is the initial horizontal position.\nThe vertical net force is -mg, so -mg=ma, thus -g = a so the projectile is accelerated by -g (the acceleration due to gravity at sea level). Integrating, we get y(t) = -(1/2)gt^2 + (v_y)t + k where v_y is the initial vertical velocity and k is the initial vertical position.\nSolving for t in the first equation, we get t(x) = (x - h)/v_x. Since the times in both equations are the same, we plug this value into y(t) to get y(t(x)) = y(x) = -(1/2)g((x - h)/v_x)^2 + (v_y/v_x)(x - h) + k.\nThus, the path of the projectile is a parabola. Setting y=0 will get you the x-intercepts of the parabola, or where the projectile touches the ground.  Note that the projectile starts above ground and time only runs in one direction, so the projectile only touches ground on one root of the equation.
714514
a) they give you everything you need.  v0 = 32, s0 = 0, so\n\ns(t) = -16t² + 32t\n\nb)\n\ns(1) = -16 + 32 = 16\n\nc)  The ball hits the ground when s(t) = 0 (the second time).\n\n0 = -16t² + 32t = -16t(t-2) so t=0 or t-2 = 0, t=2.\n\nSo the ball hits the ground when t=2.\n\nd)  The maximum height is attained at the vertex of the parabola.\n\ny - 16 = -16(t² - 2t + 1)\ny - 16 = -16(t - 1)²\n\nSo the vertex is (1, 16), that is, at time t = 1, when the height of the ball is 16.
1083521
As Ben Kenobi told Luke, "You must learn the Force."\n\nYou must learn the term Frame of Reference.\n\nIf you are on the jet doing 3000fps.  How fast are YOU moving with respect to your frame of reference.  Answer  ZERO!  Things are coming at your frame of reference at 3000fps.\n\nSo if you fired the gun, the bullet would leave the muzzle at 3000 fps from YOUR frame of reference.  From an object straight in front of the plane that you were shooting at, would see the bullet coming at them at 6000ft per second plus whatever velocity they had moving toward you.\n\nGet it?  Think from the frame of reference!\n\nYou can do it in your car.  At 60mph on the speedometer, what is your speed from your frame of reference.  Zero!  However, that possum in the middle of the road is coming at you at 60mph from YOUR frame of reference.  From the possum's frame of reference he (or she) is moving at zero.  However, the car is coming at the possum at 60mph.\n\nAs for the sound waves, the same principle holds.  They move away from you at 1100 ft/sec from YOUR frame of reference.  You would hear yourself immediately.\n\nHope this explains it.
28160
Firing at a target that is immediately in front of the weapon.  Ranges are usually measured in yards, for example 3.5yd = "three point five yards".  When you fire so close that it is immeasurable, it is 0.0yd.  People dropped the first "zero", and slang made the second zero read as "blank", so 0.0yd became "point blank" range.
1078996
When bullets are fired straight into the air they go well.....up.  Just like you and me when we jump on a trampoline eventually gravity will get the better of it and it will start coming back down.  Now if that were the only thing to take into account it would just come straight back down and thump you on the head, but because of extra forces caused by air flow and rotation the bullet won't go exactly straight up and down (thankfully).  Now a bullet that's shot straight up is going an average 900 m/s, which means it'll get up about 4,000 meters in the air and when taking into account air resistance it'll come back down with a speed of about 100 m/s.  This is a fraction of the speed but is still fast enough to pierce human flesh.  To answer the final question, the apogee is the highest point that the bullet reaches.
824387
The ball goes from 0 to 9/s in .15 secs\n\nSo:    9meters           1\n        ------------  x    ------                  =9 /.15 m/sec2\n           second          .15secs\n\n                                                      =60 m/sec 2
439275
1a)\n\n18 / 7.5 = 36 / 15 = 12 / 5\n12t - 5t^2 = t(12 - 5t) = 0 at 0, 12 / 5\n\nSo at time T = 0 the target is at the origin at the lower left hand corner, and at time T = 12 / 5 = 2.4 the target is at the lower right hand corner (18, 0).\n\n2b)\n\nAt t = 0, rM = (14, 12) is where the missile is fired from.\n\n2a)\n\nThe x coordinate of the missile is a constanr at 14, i.e., the missile is fired straight up.  So it must the target at (14, ?).\n\nThe target moves right according to 7.5t.\n\n14 / 7.5 = 28 / 15 is the time when the target would pass through x = 14.\n\ny = (12t - 5t^2) at time 28 / 15 is\n(12 * 28 / 15) - (5 * 28 * 28 / (15 * 15))\n = (1 / (15 * 15)) (12 * 15 * 28 - 5 * 28 * 28)\n = (5 * 28 / (15 * 15)) (12 * 3 - 28)\n = (5 * 8 * 28) / (15 * 15)\n = (8 * 28) / (3 * 15) = 224 / 45\n\nSo 12 - ct = 224 / 45 at time t = 28 / 15,\nc = (12 - (224 / 45)) / (28 / 15)\n = ((12 * 45) - 224) / (28 * 3)\n = (540 - 224) / (28 * 3)\n = 316 / (28 * 3) = 79 / 21\n\nwith c = 79 / 21\nat time 28 / 15\nthe target is hit at (14, 224 / 45)\n\n3)\n\nI'll let someone else graph something in proportional font ASCII where multiple whitespace is shown collapsed into a single space.
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.
956205
First compute the fall time of the keys.  Since you threw them horizontally, the fall is due only to gravity:\n\nd=1/2*g*t^2\n64=.5*9.8*t^2\nt=sqrt(2*64/9.8)\nonce you have t, just multiply the horizontal velocity times the time of flight to get the distance traveled:\n\nd=v*t\nd=8*sqrt(2*64/9.8)\n=29m\n\nIf you didn't understand how to compute the fall time, here's a bit more info:\n\nThe average velocity is\n1/2*g*t\ncompute the fall distance as\n1/2*g*t*t\n\nj
946959
Depends on the bullet and its grain.The average velocity of a 12 pdr projectile was around 900 fps with a range of \n1800 yards. ...  Field Howitzer shell
1087896
At what altitude is the lead bullet being fired??   And what is the rest of your question??