You cannot. You need to know either the initial speed or angle of projection (A).
Range of a projectileThe path of this projectile launched from a height y0 has a range d.In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range. As in Trajectory of a projectile, we will use:g: the gravitational acceleration-usually taken to be 9.80 m/s2 (32 f/s2) near the Earth's surfaceθ: the angle at which the projectile is launchedv: the velocity at which the projectile is launchedy0: the initial height of the projectiled: the total horizontal distance travelled by the projectileWhen neglecting air resistance, the range of a projectile will beIf (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify toso to increase the range θ shoud vary from 0 to 45 and after 45 it starts decreasing.
Yep that is correct. To understand this it may help you to draw a parabola. If you draw a line from the top of the parabola back to the ground you'd notice either side of this line is symmetrical. This isn't quite what happens to a projectile (following a parabolic path), but because of the nature of the question, effects such as air resistance can be ignored. As the projectile approaches the top of its path, the vertical component of its velocity approaches zero. As the projectile begins to fall the magnitude of the vertical component of the projectile begins to increase. The only force that acts on the projectile during flight is gravity which pulls it towards the earth. Since this force and the horizontal component of the projectiles velocity are at right angles to each other, the horizontal component of the velocity is unaffected during flight . This explains the symmetry of the parabola and also means the time to reach the top of path equals the time from the top of path back to the ground. The projectile will hit the ground with the same speed as it left the ground. If you draw a horizontal line through the parabola, at the two points where the line and the parabola cross, the speed of the projectile will be the same. The only change to the balls speed during the flight comes as the vertical component of its velocity tends to zero as it reaches the top of the curve and then falling back down due to gravity. I'm unsure of your physics knowledge but hopefully this doesn't confuse you. If you have learned about vectors, then this can be simply understood/explained.
All three play a part. The first applies as the arm swings and launches the projectile. The arm continues to move until it is stopped by the retaining bar and the projectile continues to move until gravity pulls back down or it impacts a structure. The second law is shown by the projectile. A big heavy rock goes slow, a small light rock flies fast. The third comes into play as the catapult launches the projectile it pushes down on the ground and slightly backward. The projectile itself uses the third law when it impacts.
If it is only 30 degrees and the system is in the ground it is probably still safe. You should have it blown out right away before it gets any colder.
In order for snow to occur, the ground temperature must be a temperature of 32 degrees or lower.
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
Assuming the angle is the angle the initial trajectory makes with the ground, that it's launched with the same speed in both cases, that it's launched from an initial height of zero, that it stops dead as soon as it touches the ground and doesn't bounce or roll, and that we can neglect air resistance (sorry for all that detail, but it does matter)... Both projectiles will end up with the same net displacement, though the 60 degree projectile will have taken a longer path to get there.
The range of a projectile on level ground, when air resistance is ignored, isd = v2*sin(2x)/g wherev is the intial velocity of the projectile,x is the angle above the horizontal at which the projectile is launchedandg is the acceleration due to the earth's gravity.This function is a maximum when x = 45 degrees and so d is smaller for other values of x.The range of a projectile on level ground, when air resistance is ignored, isd = v2*sin(2x)/g wherev is the intial velocity of the projectile,x is the angle above the horizontal at which the projectile is launchedandg is the acceleration due to the earth's gravity.This function is a maximum when x = 45 degrees and so d is smaller for other values of x.The range of a projectile on level ground, when air resistance is ignored, isd = v2*sin(2x)/g wherev is the intial velocity of the projectile,x is the angle above the horizontal at which the projectile is launchedandg is the acceleration due to the earth's gravity.This function is a maximum when x = 45 degrees and so d is smaller for other values of x.The range of a projectile on level ground, when air resistance is ignored, isd = v2*sin(2x)/g wherev is the intial velocity of the projectile,x is the angle above the horizontal at which the projectile is launchedandg is the acceleration due to the earth's gravity.This function is a maximum when x = 45 degrees and so d is smaller for other values of x.
Range of a projectileThe path of this projectile launched from a height y0 has a range d.In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range. As in Trajectory of a projectile, we will use:g: the gravitational acceleration-usually taken to be 9.80 m/s2 (32 f/s2) near the Earth's surfaceθ: the angle at which the projectile is launchedv: the velocity at which the projectile is launchedy0: the initial height of the projectiled: the total horizontal distance travelled by the projectileWhen neglecting air resistance, the range of a projectile will beIf (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify toso to increase the range θ shoud vary from 0 to 45 and after 45 it starts decreasing.
If the non-horizontal projectile is launched abovehorizontal, thenit's the second one to hit the ground, after the horizontal one.If the non-horizontal one is launched below horizontal, then it'sthe first to hit the ground, before the horizontal one.
No.
this is false. the horizontal speed of the object has absolutely nothing to do with how long it takes to fall. if you fire a bullet on level ground it will hit the ground at the same time an apple would if you drop it from the same distance above the ground at the same time. the force of gravity is the same on everything on earth.
A catapult is used to throw things. If, for example, you look at the trebuchet, you see that it has a very heavy load, heavier then the projectile, that is set up above the ground. This gives the load and consequentially the trebuchet a lot of potential energy. The projectile is launched when the load starts to fall on the ground, when it's falling, it gets kinetic energy which is given to the projectile. The projectile uses it to build up potential energy as it flies towards it's goal, and then gains back the kinetic energy as it falls.
If a projectile takes 8 seconds to reach its maximum height, it will take another 8 seconds to return to its original elevation. Presuming it is lauched from flat ground and returns to the ground, its total time in flight would be 16 seconds. If it is launched from a hill, or at a hill, more information would be needed.
if a body is thrown having initial velocity and make angle with ground this body is known as projectile and the way is calle trajectory
Wind, elevation, trajectory, projectile weight, projectile configuration, barrel length, barrel rifling, friction or resistance in the barrel, force (charge) behind the projectile. There are other enviornental elements that can affect range as well.