To find the angle that the velocity vector of the package makes with the ground, we can break down the problem into two components: the horizontal component and the vertical component of the package's velocity.
First, let's find the horizontal component of the package's velocity. Since the package is ejected horizontally from the helicopter with an initial velocity of 12.0 m/s relative to the helicopter (and in the direction opposite the helicopter's motion), the horizontal component remains constant throughout its flight. Therefore, the horizontal component of the package's velocity is also 12.0 m/s.
Next, let's find the vertical component of the package's velocity. Since the helicopter is flying at a constant altitude of 11.4 m, the package will also have a constant vertical velocity due to the effect of gravity. We can use the formula for vertical motion under constant acceleration:
v = u + at
In this case, the initial velocity (u) of the package in the vertical direction is 0 m/s, since there is no initial vertical velocity. The acceleration (a) due to gravity is -9.8 m/s² (negative because it acts downward). The time (t) is the time it takes for the package to hit the ground.
We can rearrange the equation to solve for time:
t = (v - u) / a
Since the package is falling downward, we can take the positive square root of the expression inside the parentheses.
t = √((v - u) / a)
We know that the total time it takes for the package to hit the ground is the same as the time it takes for the helicopter to cover the horizontal distance traveled. Let's call this time "T."
T = distance / speed
The distance traveled by the helicopter is the same as the horizontal distance traveled by the package. We can use the formula for distance traveled for an object with constant velocity:
distance = speed * time
Plugging in the given values, we have:
distance = 6.2 m/s * T
Since the time (T) is the same for both the horizontal and vertical motion, we can substitute the value of T into the equation to find the distance.
Now, we have the vertical velocity and the horizontal distance traveled by the package. We can find the angle that the velocity vector of the package makes with the ground using trigonometry.
tan(theta) = vertical velocity / horizontal velocity
theta = tan^(-1)(vertical velocity / horizontal velocity)
Plugging in the values we found, we can calculate the angle (theta) as seen from the ground.