Firing from a hill Firing a projectile from an elevated position increases its range. If you know the initial velocity, you should be able to use the usual formulas for displacement (distance) in the horizontal and vertical directions to determine the initial vertical position.1 When you say the initial velocity is known, I assume that includes magnitude and direction. Since velocity is a vector, you should be able to calculate the vertical and horizontal components.2 If you know the horizontal velocity and the horizontal displacement (distance traveled), you should be able to calculate the time in flight. Once you determine the time in flight, you should be able to use that value in the formula for vertical displacement to determine the initial vertical displacement. Hint: The vertical displacement of the projectile when it hits the ground is zero (assuming you have selected the origin -- the axes of the plane in which the projectile is moving -- properly). ----------- 1. d = d0 + V0t + [1/2]at2, where d0 is the initial displacement, v0 is the initial velocity, and a is acceleration. For motion in the vertical direction, a = -g. For motion in the horizontal direction, a = 0 (for projectile problems). 2. Vx = Vcos(theta); Vy = Vsin(theta), where theta is the angle of elevation. Maximum range is achieved when theta = 45 degrees. At that angle, Vx = Vy.
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PLEASE NOTE: formatting has been messed up in this so things that are supposed to be raised to a power, the number is not a superscript. This needs to be relooked at.
Maximum range is achieved when theta=45o only if the vertical displacement is zero (i.e. the projectile begins and ends at the same elevation). If launched from a certain height h, the angle for maximum range is given by
anglemax = 1/2 cos-1 [(gh)/(v^2 + gh)]
Returning to the problem, let h = launch height, R = horizontal distance from base of launch site to landing spot, V = launch speed, A = launch angle and T = time in air. The horizontal component of the launch velocity is constant since there is no acceleration in that direction. Therefore:
Vx = R/T
V cosA = R/T
Solving for T:
T = R/[V cosA]
Consider the vertical part of the problem. This solution is given for a projectile launched from an angle above the horizontal so that the initial vertical component of the velocity is positive when the acceleration due to gravity (g) is negative. Also assumed is that the launch position is above the landing position. Let the initial position be the origin.
d = do + viT + 1/2aT^2
-h = (V sinA)T - 1/2gT^2
Substituting for T:
(equation A) -h = [VR sinA ]/[V cosA] - [gR^2]/[2V^2 cos2A]
(equation B) h = -R [tanA] + [gR^2]/[2V^2cos2A]
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If you want to find the launch angle for a given height and launch speed that gives the maximum range, multiply both sides of equation A by 2cos2(theta) and rearrange to get:
(g/v^2)R^2 - (2sinAcosA)R - 2hcos2A=0
Using these trig identities:
2sinAcosA = sin2A
cos2A = 1/2 [1 + cos2A]
the equation becomes:
(g/v^2)R^2 - (sin2A)R - h[1+cos2A] = 0
Solving for R using the quadratic formula:
R(A) = v^2/(2g)[sin2A + (sin22A + (8gh/v^2)cos2A)1/2]
Find the derivative of
R(A): R'(A)=v^2/(2g)[2cos2A+1/2(sin22A+(8gh/v^2)cos2A)-1/2(4sin2Acos2A+(8gh/v^2)(-2sinAcosA))]
Set this equal to zero to find angle (A) for maximum range (R).
Rearrange and use some trig identities to get:
1/(cos22A) - v^2/(gh)(1/cos2A) - (1+v^2/(gh))=0
Use the quadratic formula to solve for 1/(cos2A):
1/(cos2A) = (1/2)[v^2/(gh) + (v^4/(g^2h^2)+4(1+v^2/(gh)))1/2] 1/(cos2A) = v^2/(gh) + 1
A = 1/2cos-1[(gh)/(v^2+gh)]
Variables:
t = time
vx = velocity in x direction
vy = velocity in y direction
v0x = initial velocity in x direction
v0y = initail velocity in y direction
v0 = initail velocity
a = acceleration
Numbers are in meters and seconds
The projectile was launched horizontally, so we know that at t = 0, vx = v0 and vy = 0. The speed in the x direction will not change during the flight; it is equal to the initial velocity.
The particle will accelerate in the y direction due to gravity. First find how long the particle takes to reach the ground, and then find out how far it traveled in the x direction during that time.
Let's go through an example problem. Say that the projectile was launched horizontally at a speed of 200 m/s and it will fall 20 meters. How far will it travel horizontally?
First we need to know how it takes the projectile to reach the ground. We can use the equation v2 = v02+2ay to find out how fast the projectile is going in the y direction when it hits the ground. a is 9.8, which is acceleration due to gravity. It is positive here because in this example we are taking the speed at the end to be positive. v0 is 0 because the projectile is not moving in the y direction at time t = 0:
v2 = 2(9.8)(20), so v = 19.80 m/s
We can then use the equation v = v0+at to find out how long it took the projectile to reach the ground (you could say that v and a are both negative, you will get the same result):
19.80 = 0 + (9.8)(t), so t = 2.02 s
Now we need to find out how far the projectile traveled in the x direction in 2.02 seconds. The speed in the x direction is 200 m/s, so multiply that by the time:
(200)(2.02) = 404 m
The projectile in this example traveled 404 meters horizontally.
Only the vertical component of motion determines how long a projectile is in the air. The equation to use is y = 1/2gt2, where y is vertical displacement in meters in a downward direction (negative), g is acceleration due to gravity, -9.8m/s2, and t is time in seconds. To solve this equation for time, do the following:
t2 = 2y/g
t = √(2y/g)
* Because both y and g are negative, when they are divided, the result is positive.
As soon as it is launched, a projectile is in free fall.
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.
True, at relatively low speeds, if the projectile was launched at maybe the speed of light, improbable as it is, this may give it enough momentum to carry on past the point were the planet's gravity has a large effect on its motion and therefore carry on moving away from the planet until stopped by another force, however, even if the initial velocity of the projectile is the same as that planets escape velocity (the speed it needs to travel at to leve the planet) this speed is not constant and the projectile will feel a very large force from the gravity of the planet and hence be brought back down to its surface
You cannot. You need to know either the initial speed or angle of projection (A).
it was launched in 1958
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)
Some of the factors that determine the movements of a projectile include: air resistance, force of gravity, initial launch velocity, the angle a projectile was launched at, and the objects initial elevation.
A launched projectile is not an example of centrifugal force. It is an example of centripetal force.
As soon as it is launched, a projectile is in free fall.
180 Degree
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.
-- In what direction should a projectile be launched in order to achieve the maximum range ? -- Toward 45 degrees above the horizontal.
The half maximum range of a projectile is launched at an angle of 15 degree
To achieve the most distance, you should set your catapult to release a projectile at 45 degrees angle relative to horizontal plane.Higher stored potential energy doesn't neccesarily mean more distance. Distance of the shot depends mainly on starting velocity, which in turn depends on rate at which catapult transfers the energy to the projectile. A limit for that depends on construction of the catapult.
Projectile motion