Question

directions: use the trig section on first page of reference table to answer the following: ① solve for both x + y components of each initial velocity vector. (a) θ = 60° vi = 40m/s (b) θ = 30° vi = 20m/s

Explanation:

Step1: Find \(v_{ix}\) for part (a)

Use the formula \(v_{ix}=v_i\cos\theta\). Given \(v_i = 40\ m/s\) and \(\theta = 60^{\circ}\), we have \(v_{ix}=40\times\cos60^{\circ}\). Since \(\cos60^{\circ}=\frac{1}{2}\), then \(v_{ix}=40\times\frac{1}{2}=20\ m/s\).

Step2: Find \(v_{iy}\) for part (a)

Use the formula \(v_{iy}=v_i\sin\theta\). Given \(v_i = 40\ m/s\) and \(\theta = 60^{\circ}\), we have \(v_{iy}=40\times\sin60^{\circ}\). Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\), then \(v_{iy}=40\times0.866 = 34.64\ m/s\).

Step3: Find \(v_{ix}\) for part (b)

Use the formula \(v_{ix}=v_i\cos\theta\). Given \(v_i = 20\ m/s\) and \(\theta = 30^{\circ}\), we have \(v_{ix}=20\times\cos30^{\circ}\). Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\), then \(v_{ix}=20\times0.866=- 17.32\ m/s\) (negative because the \(x -\)component is in the negative \(x -\)direction).

Step4: Find \(v_{iy}\) for part (b)

Use the formula \(v_{iy}=v_i\sin\theta\). Given \(v_i = 20\ m/s\) and \(\theta = 30^{\circ}\), we have \(v_{iy}=20\times\sin30^{\circ}\). Since \(\sin30^{\circ}=\frac{1}{2}\), then \(v_{iy}=20\times\frac{1}{2}=10\ m/s\).

Answer:

a. \(v_{ix}=20\ m/s\), \(v_{iy}=34.64\ m/s\)
b. \(v_{ix}=-17.32\ m/s\), \(v_{iy}=10\ m/s\)