Final answer:
The angle of projection of the projectile with the vertical is 37 degrees, determined by calculating the initial velocity components and using the inverse tangent function.
Step-by-step explanation:
The question is about determining the angle of projection of a projectile with respect to the vertical, given its position coordinates as functions of time. The position equations for the projectile are y = (4t - 2t²) meters for vertical displacement and x = (3t) meters for horizontal displacement, where t is the time in seconds.
To find the angle of projection, we need to analyze the initial velocity components. At t = 1s, the velocity components can be obtained by differentiating the position coordinates with respect to time. The vertical velocity (Vy) is dy/dt = 4 - 4t, so at t = 0s, Vy = 4 m/s. Similarly, the horizontal velocity (Vx) is dx/dt = 3 m/s, which remains constant since there is no air resistance. Thus, at t = 0s, Vx is 3 m/s.
The angle of projection θ with respect to the horizontal can be found using the tangent function, where tan(θ) = Vy/Vx. Consequently, tan(θ) = 4/3. Using the inverse tangent function, we find θ ≈ 53 degrees. But the question asks for the angle with respect to the vertical, which would be 90 degrees minus θ, which gives us 37 degrees as the angle of projection with the vertical.