Since the particle is projected horizontally, we know that the initial vertical velocity (Vy) is zero. The only velocity component acting on the particle is the horizontal velocity (Vx) of 10 m/s.
To determine the angle made by the velocity vector and the x-axis after 4 seconds, we can calculate the vertical displacement (Δy) of the particle using the equation of motion:
Δy = Vy_initial * t + (1/2) * g * t^2
where Vy_initial is the initial vertical velocity (zero in this case), t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Δy = 0 * 4 s + (1/2) * 9.8 m/s² * (4 s)^2 = 0 + (1/2) * 9.8 m/s² * 16 s² = 0 + 78.4 m = 78.4 m
After 4 seconds, the particle has fallen vertically by 78.4 meters.
Now, to find the angle made by the velocity vector and the x-axis, we can use trigonometry. The tangent of the angle (θ) is given by the ratio of the vertical displacement (Δy) to the horizontal displacement (Δx), where Δx is the product of the horizontal velocity (Vx) and the time (t):
tan(θ) = Δy / Δx = 78.4 m / (10 m/s * 4 s) = 78.4 m / 40 m = 1.96
Taking the inverse tangent (arctan) of both sides:
θ = arctan(1.96) θ ≈ 63.43 degrees
Therefore, the angle made by the velocity vector and the x-axis after 4 seconds is approximately 63.43 degrees.