A projectileis given an initial velocity of icap + 2jcap . The catesian equation of its path
V=i+2j
x=vix * time
y=yiy (time)-4.9t^2
or
x= t
y=2 t-4.9 t^2
oops, then
y=2x-4.9x^2
To find the Cartesian equation of the projectile's path, we need to analyze its motion and determine the functions that describe the x and y coordinates of the projectile as a function of time.
Let's break down the problem:
Given:
Initial velocity (vi) = icap + 2jcap
We can separate the components of the initial velocity into the x-component (vi_x) and the y-component (vi_y):
vi_x = 1
vi_y = 2
Considering that the projectile's motion is subject to gravity (in the y-direction) and assuming no air resistance, we can apply the equations of motion to find the equations for the x and y coordinates.
In the x-direction:
Since there is no horizontal acceleration, the projectile moves with a constant velocity in the x-direction. Hence, the equation becomes:
x(t) = vi_x * t
In the y-direction:
Considering the gravitational acceleration (g = 9.8 m/s^2), the equation of motion for the y-coordinate becomes:
y(t) = vi_y * t - (1/2) * g * t^2
Combining these equations, we can express the path of the projectile in terms of x and y as a function of time:
x(t) = t
y(t) = 2t - (1/2) * 9.8 * t^2
Finally, to obtain the Cartesian equation, we rewrite y in terms of x:
y(x) = 2x - (1/2) * 9.8 * x^2
The Cartesian equation of the projectile's path is y(x) = 2x - 4.9x^2.