When a particle is projected horizontally with an initial velocity and a certain height, its trajectory is affected only by the vertical motion due to gravity. In this case, we can determine the angle made by the velocity vector and the x-axis after 4 seconds by analyzing the vertical motion.
Given: Initial height (h) = 78.4 m Initial velocity in the x-direction (horizontal) = 10 m/s Time (t) = 4 s
First, we can calculate the initial vertical velocity (v_y) using the formula for vertical motion:
v_y = g * t
where g is the acceleration due to gravity (approximately 9.8 m/s²). Since the particle is projected horizontally, its initial vertical velocity is zero.
Next, we can calculate the vertical displacement (Δy) using the formula:
Δy = v_y * t + (1/2) * g * t²
Substituting the values, we have:
78.4 m = 0 * 4 s + (1/2) * 9.8 m/s² * (4 s)²
Simplifying the equation:
78.4 m = (1/2) * 9.8 m/s² * 16 s²
Solving for gives:
78.4 m = 78.4 m
The vertical displacement is equal to the initial height, which means the particle will hit the ground after 4 seconds.
Since the particle hits the ground, the angle made by the velocity vector and the x-axis after 4 seconds will be 90 degrees (or π/2 radians). This angle indicates that the particle's velocity vector is perpendicular to the x-axis at the time of impact.