Respuesta :
Answer:
vₓ = v₀ₓ , x = x₀ + v₀ₓ t
[tex]v_{y}[/tex] = v_{y} - g t, y = y₀ + v_{oy} t - ½ g t²
Explanation:
The kinematic equations in one dimension are
v = v₀ + a t
x = x₀ + v₀ t + ½ a t²
In the case of two dimensions we can separate the movement into two independent components,
X axis
In this axis there is no acceleration (a= 0), therefore the equations remain
vₓ = v₀ₓ
x = x₀ + v₀ₓ t
The subscript x is entered to indicate the direction of movement
Y Axis
In this case there is an acceleration that points down
a = - g
The equations remain
[tex]v_{y}[/tex] = The kinematic equations in one dimension are
.v = vo + a t
.x = xo + vo t + ½ to t2
In the case of two dimensions we can separate the movement into two independent components,
X axis
In this axis there is no acceleration, therefore the equations remain
.vx = vox
.x = xo + vox t
The subscript x is entered to indicate the direction of movement
Axis and In this case there is an acceleration that points down
.a = - g
The equations remain
[tex]v_{y}[/tex] = v_{y} - g t
y = y₀ + v_{oy} t - ½ g t²
The subscript and indicates the direction of movement
Answer:
For velocity;
x-component ; f(t) = V(x) = V°(x) = D(x)/t (V°(x) is considered as initial velocity along x-axis and D(x) as distance along x-axis)
y-component; f(t) = V(y) = V°(y) - gt (V°(y) is considered as initial velocity along y-axis)
For position;
x-component; f(t) = D(x) = V°(x)/t
y-component; f(t) = D(y) = V°(y) - (1/2)g(t^2)
Explanation:
In projectile motion which is a two dimensional motion, one would have to calculate the x-component and y-component separately.
Now velocity is distance/time, so it will be calculated for x and y component separately. I have considered velocity v(x) as function of time, since acceleration along x-axis in projectile motion is zero, so initial velocity V°(x) is equal to V(x), which implies that on x-axis, at any point velocity will be same. So we have to use the formula distance upon time here.
y- component of velocity is easily derived as function of time by first equation of motion which is;
V(f) = V(i) + at (since, acceleration is acting on opposite side so 'g' will be negative in value)
Now for position or distance, we also have to calculate x-component and y- component separately. x-component.
So, x-component is simply derived by multiplying time to velocity, where for y-axis, i used scond equation of motion which is;
S= V(i)t + (1/2)a(t^2) (since, acceleration is acting on opposite side so 'g' will be negative in value)
