As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of 140 m/s, and each tank is at the same height of 4.00 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from each plane. Give the directional angles with respect to the horizontal.

For plane A I got Magnitude -313.045 m/s and direction -64.41, I don't know what I'm doing wrong, I've done the problem several times. Help please!

The magnitudes will be the same. And directions. Think why.

Vf=vI+g*t
the vi is the original velocity, and the gt term is a downward term. The only difference in Vi is the vertical component, but in the upwards angle, gt will be bigger to make the result vf the same.
Magnitue:
Ke at impact=initialKE+mgh
find vf from KE.

I see how the magnitudes and directions will be the same,I just don't understand why I'm getting the numbers wrong.

To find the magnitude and direction of the velocities with which the fuel tanks hit the ground for each plane, we can use the following steps:

Step 1: Calculate the vertical component of the initial velocity for each plane.
The vertical component of velocity can be found using the formula:

v_vertical = v * sin(theta)

where v is the initial speed (140 m/s) and theta is the angle above or below the horizontal (15°).

For plane A:
v_vertical_A = 140 m/s * sin(15°)

For plane B:
v_vertical_B = 140 m/s * sin(-15°)

Note: Since the angle below the horizontal is in the opposite direction, it will have a negative sine value.

Step 2: Calculate the time of flight for each fuel tank.
The time of flight can be found using the formula:

t = 2 * v_vertical / g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

For plane A:
t_A = 2 * v_vertical_A / g

For plane B:
t_B = 2 * v_vertical_B / g

Step 3: Calculate the horizontal distance traveled by each fuel tank.
Since the initial horizontal velocity for both planes is the same, the horizontal distance can be found using the formula:

d = v * cos(theta) * t

where v is the initial speed (140 m/s), cos(theta) is the horizontal component of the initial velocity, and t is the time of flight.

For plane A:
d_A = 140 m/s * cos(15°) * t_A

For plane B:
d_B = 140 m/s * cos(-15°) * t_B

Step 4: Calculate the final velocity for each fuel tank using the Pythagorean theorem.
The final velocity can be found by combining the horizontal and vertical components of the initial velocity.

For plane A:
v_final_A = sqrt((d_A)^2 + (4.00 km)^2) / t_A

For plane B:
v_final_B = sqrt((d_B)^2 + (4.00 km)^2) / t_B

Step 5: Calculate the magnitude and direction of the final velocity for each fuel tank.
The magnitude of the final velocity can be found using the formula:

magnitude = |v_final|

The direction of the final velocity can be found using the inverse tangent function:

direction = atan((4.00 km) / d)

where d is the horizontal distance traveled.

For plane A:
magnitude_A = |v_final_A|
direction_A = atan((4.00 km) / d_A)

For plane B:
magnitude_B = |v_final_B|
direction_B = atan((4.00 km) / d_B)

Now, follow these steps with the given values (140 m/s, 15°, -15°) to find the correct results for plane A and plane B.

To find the magnitude and direction of the velocity with which the fuel tank hits the ground, you need to use the concepts of projectile motion. Here's a step-by-step guide to solve the problem:

Step 1: Identify the given information and the parameters you're trying to find. In this case, the given information is:
- Initial speed of both planes: 140 m/s
- Height of the tanks above the ground: 4.00 km
- Angle of plane A (above horizontal): 15.0°
- Angle of plane B (below horizontal): -15.0°

You're trying to find the magnitude and direction of the velocity with which the fuel tank hits the ground for each plane.

Step 2: Use the given information and the equations of projectile motion to solve the problem. Recall the following equations:
- Vertical displacement, Δy = (viy × t) + (0.5 × a × t^2)
- Vertical velocity, vfy = viy + (a × t)
- Horizontal displacement, Δx = vix × t
- Horizontal velocity, vfx = vix

Step 3: First, calculate the time taken to reach the ground for each tank by using the vertical displacement formula. Since both tanks are at the same height of 4.00 km, the vertical displacement is the same for both.

Δy = (viy × t) + (0.5 × a × t^2)
0 = (viy × t) + (0.5 × (-9.8 m/s^2) × t^2) (taking downward acceleration as negative)
0 = (viy × t) - (4.9 m/s^2) × t^2

Solve this quadratic equation to find the time t. You will get two solutions, one positive and one negative. Since time cannot be negative, consider the positive value.

Step 4: Once you have the time t, you can find the horizontal displacement, Δx, for each tank by using the horizontal displacement formula.

Δx = vix × t

Step 5: Finally, you can find the magnitude and direction of the velocity with which the fuel tank hits the ground. The magnitude can be calculated using the Pythagorean theorem:

Magnitude = sqrt((Δx)^2 + (Δy)^2)

The direction can be found using the arctan function:

Direction = arctan(Δy / Δx)

Now, plug in the values for plane A and plane B separately, and calculate the magnitude and direction for each.

If you're still having trouble or getting incorrect answers, please provide your work, and I can help you identify any mistakes.