An airplane with a speed of 85.1 m/s is climbing upward at an angle of 54.4 ° with respect to the horizontal. When the plane's altitude is 759 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 separate it into vertical and horizontal components. Let's start with the vertical component.

(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth:

1. To find the time it takes for the package to hit the ground, we need to consider the vertical motion. We can use the formula:
h = v₀t + (1/2)gt²,
where h is the altitude (759 m), v₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s²).

2. The initial vertical velocity can be obtained from the given speed and angle. Using trigonometry, we have:
v₀ = v × sin(θ),
where v is the speed of the airplane (85.1 m/s) and θ is the angle with respect to the horizontal (54.4°).

Therefore, v₀ = 85.1 m/s × sin(54.4°).

3. Substitute this value into the equation from step 1:
759 m = (85.1 m/s × sin(54.4°))t + (1/2) × (-9.8 m/s²)t².

4. Rearrange the equation to find the time, t:
(1/2)(-9.8) t² + (85.1sin(54.4))t - 759 = 0.

5. Solve the quadratic equation using the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / (2a),
where a = (1/2)(-9.8), b = 85.1sin(54.4), and c = -759.

6. Calculate the two possible values of t, and select the positive value since time cannot be negative.

7. Once you have the value of t, substitute it back into the equation from step 3 to find the horizontal distance traveled:
distance = v × cos(θ) × t,
where θ is the angle with respect to the horizontal (54.4°).

Therefore, distance = 85.1 m/s × cos(54.4°) × t.

(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact:

1. To find the velocity vector just before impact, we can use the horizontal distance found in part (a) and the time, t.

2. The horizontal velocity of the package remains constant throughout the motion, as there are no horizontal forces acting on it. Therefore, the horizontal component of the velocity just before impact is the same as the initial horizontal velocity of the airplane, which is v × cos(θ).

3. The vertical component of the velocity just before impact can be calculated using the equation:
v' = v₀ + gt,
where v₀ is the initial vertical velocity (obtained from v × sin(θ)) and g is the acceleration due to gravity (-9.8 m/s²).

4. Calculate the angle of the velocity vector using the horizontal and vertical components of the velocity just before impact.

And there you have it! You can now calculate the distance along the ground to where the package hits the earth and 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:

(a) Finding the horizontal distance traveled by the package before hitting the earth.

To calculate the horizontal distance, we need to find the time it takes for the package to reach the ground. Assume that the time of flight is "t" seconds.

To find 't,' we can use the equation of motion in the vertical direction:

y = yo + vot + (1/2)at^2

where y is the final vertical position of the package (0 m at the ground), yo is the initial vertical position (759 m), vo is the initial vertical velocity, a is the vertical acceleration (-9.8 m/s^2, due to gravity), and t is the time of flight.

Since the package is launched horizontally, vo in the vertical direction is 0 m/s. Therefore, the equation simplifies to:

0 = 759 + 0 + (1/2)(-9.8)t^2

Simplifying further:

0 = 759 - 4.9t^2

Rearranging the equation gives:

t^2 = 759 / 4.9

t ≈ 7.18 seconds

Now that we know the time of flight, we can calculate the horizontal distance traveled by the package using the equation:

d = v * t

where d is the horizontal distance, v is the horizontal velocity, and t is the time of flight.

In this case, the horizontal velocity (v) is given as 85.1 m/s. Substituting the calculated value of 't':

d = 85.1 * 7.18

d ≈ 611.5 meters

Therefore, the package will hit the earth approximately 611.5 meters away from the point directly beneath where it was released.

(b) Finding the angle of the velocity vector of the package just before impact.

To find the angle, we can use trigonometry. The horizontal component of the velocity will remain constant at 85.1 m/s, and the vertical component of the velocity can be calculated using the equation:

v = vo + at

In this case, vo in the vertical direction is 0 m/s, and the acceleration is -9.8 m/s^2. Substituting these values:

v = 0 + (-9.8) * 7.18

v ≈ -70.5 m/s (negative because the velocity is downward)

Now, using the equation:

θ = arctan(v_horizontal / v_vertical)

θ = arctan(85.1 / -70.5)

θ ≈ -50.5°

Therefore, the angle of the velocity vector of the package just before impact, relative to the ground, is approximately -50.5°. Note that the negative sign indicates a downward direction.