An airplane with a speed of 82.2 m/s is climbing upward at an angle of 71.6 ° with respect to the horizontal. When the plane's altitude is 772 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion of the airplane and the package.

(a) To calculate the distance along the ground to where the package hits the Earth, we need to find the horizontal displacement of the package.

First, let's find the time it takes for the package to hit the ground. Since we know the initial altitude of the plane (772 m), we can use the equation of motion:

Δy = v₀y * t + (1/2) * a * t²

Where:
Δy = final altitude - initial altitude = 0 - 772 m
v₀y = initial vertical velocity = v * sin(θ) = 82.2 m/s * sin(71.6°)
a = acceleration due to gravity = -9.8 m/s² (since it works opposite to the defined positive direction)
t = time

Rearranging the equation, we get:

0 = (82.2 m/s * sin(71.6°)) * t + (1/2) * (-9.8 m/s²) * t²

Now, we can solve this equation for t. Plugging the given values into the equation:

0 = (82.2 m/s * sin(71.6°)) * t - 4.9 m/s² * t²

Simplifying the equation:

-4.9 t² + (82.2 * sin(71.6°)) t = 0

By factoring out t, we get:

t * (-4.9 t + (82.2 m/s * sin(71.6°))) = 0

Since t cannot be zero (as it represents the time taken for the package to hit the ground), we solve:

-4.9 t + (82.2 m/s * sin(71.6°)) = 0

Solving for t, we get:

t = (82.2 m/s * sin(71.6°)) / 4.9 m/s²

Now, we have found the time taken for the package to hit the ground.

Next, we can calculate the horizontal displacement or the distance along the ground using the formula:

Δx = v₀x * t

Where:
Δx = distance along the ground
v₀x = initial horizontal velocity = v * cos(θ) = 82.2 m/s * cos(71.6°)
t = time taken to hit the ground (calculated above)

Plugging in the values:

Δx = (82.2 m/s * cos(71.6°)) * (v₀y / a)

Calculating this expression will give us the distance along the ground where the package hits the Earth.

(b) To determine the angle of the velocity vector of the package just before impact, we can use trigonometry.

The vertical component of the velocity (v₁y) just before impact can be calculated by subtracting the time taken to hit the ground (t) from the initial vertical velocity (v₀y):

v₁y = v₀y - a * t

Using the calculated values of v₀y and t, we can find v₁y.

The horizontal component of the velocity (v₁x) just before impact remains constant and is equal to the initial horizontal velocity (v₀x):

v₁x = v₀x

Now, we can calculate the magnitude of the velocity just before impact (v₁) using the Pythagorean theorem:

v₁ = √(v₁x² + v₁y²)

Finally, we can find the angle of the velocity vector just before impact (θ₁) using inverse trigonometric functions:

θ₁ = atan(v₁y / v₁x)

Solving this equation will give us the angle of the velocity vector just before impact relative to the ground.