A projectile is given an initial velocity of (i + 2j) m/s, where i is along the ground and j is along the vertical. If g = 10 m/s2, the equation of its trajectory is:
To find the equation of the trajectory of a projectile, we first need to separate the horizontal and vertical components of its motion.
Let's consider the horizontal and vertical components of the initial velocity separately:
Horizontal component: The initial velocity is given as (i + 2j) m/s. The "i" component represents the motion along the ground, so the horizontal component is the coefficient of "i," which is 1 m/s.
Vertical component: The vertical component is the coefficient of "j," which is 2 m/s.
Now, to determine the equation of the trajectory, we need to consider the equations of motion for horizontal and vertical motion independently.
Horizontal motion:
The horizontal motion of a projectile is constant unless there is air resistance or external forces acting on it. Therefore, the equation for horizontal motion is given by:
x = (initial horizontal velocity) * t
Vertical motion:
The vertical motion of a projectile is influenced by the acceleration due to gravity. The equation for the vertical motion of a projectile is given by:
y = (initial vertical velocity) * t - 0.5 * g * t^2
Now, substituting the values of initial vertical velocity, initial horizontal velocity, and acceleration due to gravity into the equations, we get:
x = (1 m/s) * t
y = (2 m/s) * t - 0.5 * (10 m/s^2) * t^2
So, the equation of the trajectory of the projectile is given by:
x = t
y = 2t - 5t^2
Note: In this case, we have assumed that the projectile is starting from the origin (0,0) and neglecting any initial height above the ground.
v = 2 - 10 t
y = 2 t - 5 t^2
x = 1 t
so t = x
y = 2 x - 10 x^2