To determine the velocity of the body after 16 seconds, we need to break down the initial velocity into its horizontal and vertical components.
Given: Initial velocity (v0) = 200 m/s Launch angle (θ) = 30 degrees Time (t) = 16 seconds
To find the horizontal and vertical components of the velocity, we can use the following equations:
Horizontal component: v0x = v0 * cos(θ) Vertical component: v0y = v0 * sin(θ)
Substituting the given values:
v0x = 200 * cos(30°) v0y = 200 * sin(30°)
v0x ≈ 173.21 m/s v0y ≈ 100 m/s
Now, let's analyze the vertical motion to determine the final vertical velocity after 16 seconds. We'll use the equation:
v = v0y + at
where: v = final vertical velocity v0y = initial vertical velocity a = acceleration (due to gravity, approximately -9.8 m/s^2) t = time
Substituting the known values:
v = 100 - 9.8 * 16 v ≈ -36.8 m/s (negative sign indicates downward direction)
The vertical velocity after 16 seconds is approximately -36.8 m/s.
Since the horizontal component of velocity (v0x) remains constant, the horizontal velocity after 16 seconds remains the same:
v0x ≈ 173.21 m/s
Therefore, the velocity of the body after 16 seconds can be represented as a vector:
Velocity = v0x î + v î ≈ 173.21 m/s î - 36.8 m/s ĵ
where î and ĵ are unit vectors along the x-axis and y-axis, respectively.