Need derivation of projectile motion

[piyush3dxyz]

Hey i need derivation of projectile motion..
All four equation..
including range,height,X-axis distance,y-axis distance..

 

[Doc Al]

Try this: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html"

 

[piyush3dxyz]

Try this: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html"

thx a lot sir...
regards piyush...

 

[BruceW]

A projectile has constant acceleration downwards of g (roughly equal to 9.81m/s^2). And it has zero acceleration horizontally.

Since the acceleration is completely specified, we can use integration to find the laws of motion. Using z=upward and x being horizontal:

$\ddot{z}=-g$ and $\ddot{x}=0$. Now using integration on both equations individually:
$\dot{z} = \dot{z_0} -g(t - t_0)$ and $\dot{x} = \dot{x_0}$ (where the subscript zero are constants - i.e. boundary conditions). So now we integrate the equations again to get:
$z = z_0 + \dot{z_0}t -\frac{1}{2}g(t - t_0)^2$ and $x = x_0 +\dot{x_0}t$

These are all the laws of motion necessary. If you get a question that tells you the initial angle an object is fired at, then that gives you the ratio of the horizontal and vertical initial speeds. And if you get a question asking where an object ends up landing, then often you can solve for the vertical motion to get the time, and then use the time in the horizontal motion equation to get where it ends up landing.

 

[PJC01]

Sure, I'd be happy to provide a derivation of projectile motion and the four equations associated with it. Projectile motion is the motion of an object that is launched into the air and then moves along a curved path under the influence of gravity. The four equations of motion for projectile motion are:

1. Range (R) = (v^2 * sin2θ)/g
2. Maximum height (H) = (v^2 * sin^2θ)/(2g)
3. Time of flight (t) = 2v * sinθ/g
4. Horizontal distance (x) = v * cosθ * t

Where:
v = initial velocity of the object
θ = angle of launch
g = acceleration due to gravity (9.8 m/s^2)

Now, let's derive these equations step by step:

1. Range (R): To derive the range equation, we first need to consider the horizontal and vertical components of the object's motion separately. The horizontal component of the object's velocity remains constant throughout the motion, while the vertical component is affected by gravity.

Horizontal component: The horizontal velocity (vx) remains constant throughout the motion and can be represented as:
vx = v * cosθ

Vertical component: The vertical velocity (vy) changes due to the acceleration of gravity and can be represented as:
vy = v * sinθ - gt

The time taken for the object to reach the ground is equal to the time taken for the object to reach its maximum height (H). Therefore, we can equate the two equations for vy and solve for t:
v * sinθ - gt = 0
t = 2v * sinθ/g

Now, we can substitute this value of t into the horizontal equation to find the range (R):
x = vx * t
x = (v * cosθ) * (2v * sinθ/g)
x = (v^2 * sin2θ)/g

2. Maximum height (H): To derive the maximum height equation, we can use the same equation for vy and set it equal to zero (since the object reaches its maximum height when vy = 0):
v * sinθ - gt = 0
v * sinθ = gt
H = vy * t - (1/2)gt^2
H = (v * sinθ) * (2v * sinθ/g) - (1/2)g(