Let's answer **questions 5, 6, and 7** step by step: --- ## 5. Algebraically solve the relations for the components of the initial velocity vector (**v₀ₓ**, **v₀ᵧ**). **Given:** From standard projectile motion, the position as a function of time is: - \( x(t) = v_{x} t \) - \( y(t) = v_{y} t - \frac{1}{2} g t^2 \) Suppose you are given the horizontal distance \( d \) and vertical distance \( h \) that the projectile travels in time \( t \). ### **Step 1: Express \( v_{x} \) and \( v_{y} \) in terms of \( d, h, t, g \):** \[ x(t) = v_{x} t \implies v_{x} = \frac{d}{t} \] \[ y(t) = v_{y} t - \frac{1}{2} g t^2 \implies v_{y} = \frac{h + \frac{1}{2} g t^2}{t} \] --- ## 6. x- and y-components of the initial velocity in terms of magnitude and angle Let the initial velocity magnitude be \( v_ \) and the angle with the horizontal be \( \theta \): \[ v_{x} = v_ \cos\theta \] \[ v_{y} = v_ \sin\theta \] --- ## 7. Position as a function of time Substitute the expressions from above into the position equations: \[ x(t) = v_{x} t = \left(\frac{d}{t}\right)t = d \] \[ y(t) = v_{y} t - \frac{1}{2} g t^2 = \left(\frac{h + \frac{1}{2} g t^2}{t}\right)t - \frac{1}{2} g t^2 = h + \frac{1}{2} g t^2 - \frac{1}{2} g t^2 = h \] This confirms the initial values at \( t \). If you want the general expressions for any time \( t \), use: \[ x(t) = v_{x} t = \frac{d}{t_f} t \] \[ y(t) = v_{y} t - \frac{1}{2} g t^2 = \frac{h + \frac{1}{2} g t_f^2}{t_f} t - \frac{1}{2} g t^2 \] Where \( t_f \) is the total flight time. --- ### **Summary Table** | Step | Expression | |------|------------| | 5 | \( v_{x} = \frac{d}{t} \), \( v_{y} = \frac{h + \frac{1}{2} g t^2}{t} \) | | 6 | \( v_{x} = v_ \cos\theta \), \( v_{y} = v_ \sin\theta \) | | 7 | \( x(t) = \frac{d}{t_f} t \), \( y(t) = \frac{h + \frac{1}{2} g t_f^2}{t_f} t - \frac{1}{2} g t^2 \) | --- Let me know if you need numbers plugged in or further explanation!