Horizontal Motion: X₁ = xo +Vxot Vxf = Vxo Vertical Motion: Eqn. A y, = y + vt + ½a,t² Vyg=vyo+a,t 2 2 Eqn. B Eqn. C +2a, (v, - y) Eqn. D where Vyo and are the initial vertical and horizontal components of the velocity respectively, x, and x are the final and initial horizontal positions, y, and y, are the final and initial vertical positions. 1. Notice that Equations A and B have a common variable, t. Equation A predicts the x coordinate in terms of the parameter t, Equation B predicts the y coordinate in terms of the parameter, t. Combine these two equations to remove the t variable from the equation. Solve Equation A for t and substituting it into Equation B: What equation for y, do you get? The components of the velocity can be written in terms of the original launch velocity as: Vo=V₁ cose Vov, sine The components of x-x, can be written as Ax. 2. These components, when combined with the equation you determined above (#1) yield an equation for y(x) determined completely by vo and ] (the initial launch speed and angle). Write this equation below:

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Work on 1 and 2 because they are both related to each other
### Activity 1: Calculation of projectile equation 

**The purpose of this activity is to calculate the equation for the final horizontal position of a projectile using only the knowledge of its initial height, angle of projection, and acceleration.**

#### Horizontal Motion:
\[ x_f = x_0 + v_{x0}t \quad \text{Eqn. A} \]
\[ v_{x_f} = v_{x0} \]

#### Vertical Motion:
\[ y_f = y_0 + v_{y0}t + \frac{1}{2}a_y t^2 \quad \text{Eqn. B} \]
\[ v_{y_f} = v_{y0} + a_y t \quad \text{Eqn. C} \]
\[ v_{y_f}^2 = v_{y0}^2 + 2a_y (y_f - y_0) \quad \text{Eqn. D} \]

where \( v_{y0} \) and \( v_{x0} \) are the initial vertical and horizontal components of the velocity respectively, \( x_f \) and \( x_0 \) are the final and initial horizontal positions, \( y_f \) and \( y_0 \) are the final and initial vertical positions.

1. Notice that Equations A and B have a common variable, \( t \). Equation A predicts the \( x \) coordinate in terms of the parameter \( t \), and Equation B predicts the \( y \) coordinate in terms of the parameter \( t \). Combine these two equations to remove the \( t \) variable from the equation. Solve Equation A for \( t \) and substitute it into Equation B: What equation for \( y_f \) do you get?

The components of the velocity can be written in terms of the original launch velocity as:
\[ v_{x0} = v_0 \cos \theta \]
\[ v_{y0} = v_0 \sin \theta \]

The components of \( x_f - x_0 \) can be written as \( \Delta x \).

2. These components, when combined with the equation you determined above (#1) yield an equation for \( y(x) \) determined completely by \( v_0 \) and \( \theta \) (the initial launch speed and angle). Write this equation below:

\[ \quad \]
Transcribed Image Text:### Activity 1: Calculation of projectile equation **The purpose of this activity is to calculate the equation for the final horizontal position of a projectile using only the knowledge of its initial height, angle of projection, and acceleration.** #### Horizontal Motion: \[ x_f = x_0 + v_{x0}t \quad \text{Eqn. A} \] \[ v_{x_f} = v_{x0} \] #### Vertical Motion: \[ y_f = y_0 + v_{y0}t + \frac{1}{2}a_y t^2 \quad \text{Eqn. B} \] \[ v_{y_f} = v_{y0} + a_y t \quad \text{Eqn. C} \] \[ v_{y_f}^2 = v_{y0}^2 + 2a_y (y_f - y_0) \quad \text{Eqn. D} \] where \( v_{y0} \) and \( v_{x0} \) are the initial vertical and horizontal components of the velocity respectively, \( x_f \) and \( x_0 \) are the final and initial horizontal positions, \( y_f \) and \( y_0 \) are the final and initial vertical positions. 1. Notice that Equations A and B have a common variable, \( t \). Equation A predicts the \( x \) coordinate in terms of the parameter \( t \), and Equation B predicts the \( y \) coordinate in terms of the parameter \( t \). Combine these two equations to remove the \( t \) variable from the equation. Solve Equation A for \( t \) and substitute it into Equation B: What equation for \( y_f \) do you get? The components of the velocity can be written in terms of the original launch velocity as: \[ v_{x0} = v_0 \cos \theta \] \[ v_{y0} = v_0 \sin \theta \] The components of \( x_f - x_0 \) can be written as \( \Delta x \). 2. These components, when combined with the equation you determined above (#1) yield an equation for \( y(x) \) determined completely by \( v_0 \) and \( \theta \) (the initial launch speed and angle). Write this equation below: \[ \quad \]
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