To determine the angle made by the velocity vector and the x-axis after 4 seconds, we need to consider the horizontal and vertical components of the particle's motion.
Given: Initial velocity (u) = 10 m/s Time (t) = 4 seconds Height (h) = 78.4 m
First, we can find the horizontal distance traveled by the particle using the horizontal component of the motion:
Horizontal distance (x) = Horizontal velocity (Vx) * Time (t)
Since the particle is projected horizontally, the initial horizontal velocity (Vx) remains constant throughout the motion. Therefore, Vx = u = 10 m/s.
x = Vx * t x = 10 m/s * 4 s x = 40 m
Next, we can calculate the time taken by the particle to reach the ground from the given height using the vertical component of the motion:
h = Vertical velocity (Vy) * Time (t) - 0.5 * Acceleration due to gravity (g) * Time (t)^2
Since the particle is projected horizontally, there is no initial vertical velocity (Vy) and the vertical distance (h) is equal to the initial height (78.4 m). Also, the acceleration due to gravity (g) is approximately 9.8 m/s^2.
78.4 m = 0 * t - 0.5 * 9.8 m/s^2 * t^2 78.4 m = -4.9 m/s^2 * t^2
Solving for time (t):
t^2 = 78.4 m / (4.9 m/s^2) t^2 = 16 t = 4 seconds
The time taken to reach the ground is 4 seconds, which matches the given value.
Now, we have the horizontal distance (x = 40 m) and the vertical distance (h = 78.4 m), and we can calculate the angle using the tangent function:
Angle (θ) = arctan(h / x)
θ = arctan(78.4 m / 40 m) θ ≈ 63.4 degrees
Therefore, after 4 seconds, the angle made by the velocity vector and the x-axis is approximately 63.4 degrees.