A projectile is given an initial velocity of I +2j where I is along the ground and j is along the vertical if g =10m/s², the equation of its trajectory is

Explanation

When a projectile is launched at an angle with an initial velocity of u m/s, its horizontal and vertical components are given by:Horizontal component: u_x = u * cos(θ)Vertical component: u_y = u * sin(θ)where θ is the launch angle.In this specific case, the initial velocity is given as i + 2j, where i represents the horizontal component and j represents the vertical component. Since i corresponds to the horizontal direction and j corresponds to the vertical direction, we can directly write the components as:Horizontal component: u_x = 1 m/sVertical component: u_y = 2 m/sThe acceleration due to gravity (g) is acting in the negative vertical direction, which is represented by -10 m/s².Here's how to find the equation of the projectile's trajectory:Independent motion: We can analyze the horizontal and vertical motions of the projectile separately as they are independent of each other due to the principle of superposition.Horizontal motion: The horizontal motion is a constant velocity motion because there's no horizontal acceleration. Therefore, the horizontal position of the projectile at any time t is given by:x = u_x * t = 1 tVertical motion: The vertical motion is under the influence of constant acceleration due to gravity (-g). We can use the following equation to find the vertical position of the projectile at any time t:y = u_y * t - 1/2 * g * t²Substituting the values of u_y and g, we get:y = 2t - 1/2 * 10 * t² = 2t - 5t²Combining horizontal and vertical motion: To eliminate time t and obtain the relationship between the horizontal position x and the vertical position y, we can substitute the expression for t from the horizontal motion equation into the vertical motion equation:y = 2 (x / 1) - 5 (x / 1)²Simplifying the equation, we get:y = 2x - 5x²Therefore, the equation of the projectile's trajectory is y = 2x - 5x²