A rocket was fired at angle of 45 degree with a velocity of 600m&#92;s trace its tragectory & compute the range.
439425
You need to give more info, like fuel capacity, weight of rocket, length of fueled burst.
486759
You're trying to find the vertical angle, angle of site and site.  (But your overall goal is to find the elevation of the guns)\n\nRange and charge
519171
Also launch at 45% in the direction of the wind, less height but more distance
962545
9 hours, 55 minutes, 29.685 seconds at a rotational velocity of 45,300 km/h (28,152 mph)
496012
javelin thrown at angles 30 and 60 will land in same spot because as per properties of projection if x is angle of projection then range for angles of projection x and 90-x will be same
892312
Hmm-m. Interesting question...\n\nI thought about this for a while and see no\nreason why this math would not be correct.\nAssuming that the bullet did explode when\nstruck by the firing pin, and the bullet left \nthe end of the barrel of the gun at say \n2000 feet per second velocity...\n\n2000 fps = 1800 Mph\n\nThe Moon is 239,000 miles from Earth.\nLet's say you were 10 miles up in the sky\nwhen you fired the gun.  Distance = 238,990 Miles.\n\nSo, 238,990 / 1800 = 132.7 hours \n\nSince there is no atmosphere in space to slow down the \nbullet as it traveled, we can estimate that the bullet would \ncontinue traveling at 1800 mph for the entire trip. The \ngravitational pull of the Moon would influence the bullet's\nvelocity somewhat towards the end of its flight, but my\nfigures show that with 1/6 the gravity of Earth, the Moon\nwould not significantly change the time of flight.\n\nAnswer: 132 Hours.   or   5.5 Days.\n\nNow, I have two questions for you:\n\n#1.)  If the rifle did fire off in space, would you see any smoke?\n\n#2.)  If you were 10 miles up in the atmosphere - say 50,000 ft.,\nhow fast must you be going, to be able to remain at that altitude,\nand in which direction?\n\nYour question makes reference to the Space Station as the point of firing the gun. In my answer above I did not use that as the platform from which the gun was fired because I do not have at my finger tips the approximate altitude of the Space Station though I know it is much more than 50,000 Ft. I also did not take into account the thin atmosphere which exists at 50,000 feet, or the pull backwards on the bullet by the gravity of Earth. These omissions were made simply to run the numbers and see what I came up with since I thought this was an interesting question. Now, having dispensed with the mundane aspects of my quick math, let's discuss the real problem with this question...\n\nThe Earth is spinning. The Moon is orbiting the Earth at a certain velocity, hence it is a moving target. The moving target is being shot at from a moving platform. If you have any hope of hitting the Moon with your bullet, you would have to fire at the place where the Moon will be 5.5 days from the time of firing your gun, not at where the Moon is now. To do that requires more math than a few simple divisions of distance and velocity. For example, the moon swings completely around the Earth at a radius of 238,900 miles in 27.3 Earth days. The circumfrence of that circle is Pi x D = 1,500,292 miles (assuming the orbit is a circle, which it is not). So, the velocity that the Moon is moving away from its present spot right this minute is 2,290 miles per hour.  Since the Moon rises and sets on our horizon each night, you can see that in our calculation, the Moon would have passed overhead at least 5 times in 5.5 days. So your rifle would have to be pointed at the place you calculate the Moon should be at in 5.5 days.        \nRegards,\nZah
72528
TOW rocket launcher (Tube launched Optically sighted Weapon)\n\nStinger Missile\n.50 Caliber machine gun\nM16A2 rifle\nM1911 semi-automatic pistol\netc etc
553692
Were the rocket is before it is launched.
815666
The information give with your distances and coordinates is very unclear.\nIf you give a North and West direction for the bullet, then the angle in degrees and distance can be calculated.  And likewise for the coyote.\nYou have given two angles for one location.  Confusion!!!!!!!!
104839
When an object is launched under the influence of only a uniform gravitational force field, it is said to follow a projectile motion. Under ideal condition, the path it traces is parabolic. There is catch here – if the initial velocity is equal or more than the escape velocity then the object will simply vanish into space i.e. leave earth!\n\nImagine if the launch angle from the horizontal is “theta” and initial velocity is “U” then you can easily compute vertical (Uv) and horizontal (Uh) components of it. The Uv will be equal to U x Sin(theta) and Uh will be equal to U x Cos(theta). Now by using Uv you can compute the max height it will reach and the total time it will take to go up and return to ground. With this round trip time in hand, you can easily compute the max horizontal distance it will cover before hitting the ground.\n\nAll these calculation assume ideal conditions like no wind resistance etc.
672620
2. An oxygen molecule moving at 560 meters/second.
813272
no its wrong\nthe correct answer is \n15 degrees and 75 degrees\nbecause the range of a projectile is same at \nangle A (say) and 90 degrees- angle A
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.
164737
if it can travel at the same speed at a 45 degree angle, they will both travel the same distance.
42921
optimum angle for distance is 45 degrees \nspeed: fucking fast
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.
641828
500ft\nbut you still have to be careful of whats behind your target.