Question

101 Range Versus Height We say that an object undergoes free fall when the acceleration is only due to gravity. An object that is in free fall experiences a constant gravitational acceleration 푔 that points towards the ground. The object may travel along a complicated two-dimensional path but its motion can be decomposed into two independent one-dimensional motions. Gravity acts only in the vertical (y) direction and therefore there is no acceleration in the horizontal (x) direction. Using this information, we can write expressions for the position and velocity of the object as a function of t

Answer 1

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)

Answer 2

Answer:

vₓ = v₀ₓ ,                 x = x₀ + v₀ₓ t

$v_{y}$ = 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

           $v_{y}$ = 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

           $v_{y}$ = v_{y} - g t

           y = y₀ + v_{oy}  t - ½ g t²

The subscript and indicates the direction of movement