Part 1 of 6 A projectile is launched at an angle of \( 45^{\circ} \) to the horizontal from 6 ft above the ground at an initial speed of \( 200 \mathrm{ft} / \mathrm{sec} \). Assume the x -axis is horizontal, the positive \( y \)-axis is vertical (opposite g ), the ground is horizontal, and only the gravitational force acts on the object. Answer parts a through d . a. Find the velocity and position vectors for \( \mathrm{t} \geq 0 \). The velocity vector is \( \mathrm{v}(\mathrm{t})=\langle\square \).

We begin with the given information: • Launch angle: 45° • Initial speed: 200 ft/sec • Initial position: (x, y) = (0, 6) ft • Gravitational acceleration: g = 32 ft/sec² (acting downward) Step 1. Resolve the initial velocity into horizontal and vertical components. Since the angle is 45°:   v₀ₓ = 200 cos(45°) = 200·(√2/2) = 100√2 ft/sec   v₀ᵧ = 200 sin(45°) = 200·(√2/2) = 100√2 ft/sec Step 2. Write the velocity vector v(t) for t ≥ 0. The horizontal motion has no acceleration, so:   vₓ(t) = 100√2 ft/sec The vertical motion is under constant acceleration –g = –32 ft/sec², so:   vᵧ(t) = 100√2 – 32t ft/sec Thus, the velocity vector is:   v(t) = ⟨100√2, 100√2 – 32t⟩ Step 3. Now, integrate the components to find the position vector r(t). For the horizontal component:   x(t) = ∫(100√2) dt = 100√2 · t + Cₓ Since x(0) = 0, we have Cₓ = 0. Thus, x(t) = 100√2 · t For the vertical component:   y(t) = ∫(100√2 – 32t) dt = 100√2 · t – 16t² + Cᵧ With y(0) = 6, we get Cᵧ = 6. Thus, y(t) = 6 + 100√2 · t – 16t² So, the position vector is:   r(t) = ⟨100√2 · t, 6 + 100√2 · t – 16t²⟩ Answer Summary: a. The velocity vector for t ≥ 0 is:   v(t) = ⟨100√2, 100√2 – 32t⟩ and the position vector is:   r(t) = ⟨100√2 · t, 6 + 100√2 · t – 16t²⟩