A helicopter is flying in a straight line over a level field at a constant speed of 6.2 m/s and at a constant altitude of 11.4 m. A package is ejected horizontally from the helicopter with an initial velocity of 12.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion. What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

The following calculations assume no air resistance.

Calculate the vertical component of velocity v at impact from the vertical distance S=-11.4 m, a=-g=-9.8 m/s/s, and initial velocity u=0 (package ejected horizontally) from
v²-u²=2aS

The horizontal component of velocity (h) at impact is the same as at ejection, namely 6.2-12=5.8.

The angle with the horizontal is tan-1(v/h)
=tan-1(v/5.8)

To find the angle the velocity vector of the package makes with the ground at the instant before impact, we need to use trigonometry.

Let's break down the horizontal and vertical components of the package's velocity relative to the ground:

Horizontal component: The package is ejected horizontally from the helicopter, so its horizontal velocity relative to the ground will be the same as the helicopter's velocity, 6.2 m/s.

Vertical component: The package is ejected with an initial velocity of 12.0 m/s relative to the helicopter. Since the helicopter is flying at a constant altitude, the package's vertical velocity relative to the ground will be the same as the velocity at which it was ejected vertically from the helicopter, which is 12.0 m/s.

Now, we can calculate the angle the velocity vector of the package makes with the ground using the tangent function:

tan(theta) = vertical component / horizontal component
tan(theta) = 12.0 m/s / 6.2 m/s

Taking the inverse tangent of both sides, we have:

theta = arctan(12.0 m/s / 6.2 m/s)

Using a calculator, we find:

theta ≈ 63.6 degrees

Therefore, the angle the velocity vector of the package makes with the ground at the instant before impact, as seen from the ground, is approximately 63.6 degrees.

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.