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Q: How do you calculate range of projectile from ground at A degrees if you know its range is 200 meters if its launched from height of 2 meters at A degrees?

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For a given velocity, the maximum distance will be achieved when the projectile is launched at an angle of 45o (neglecting air resistance).

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.

No.

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 surfaceÎ¸: 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 Î¸ shoud vary from 0 to 45 and after 45 it starts decreasing.

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.

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.

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

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